2017"百度之星"程序设计大赛-资格赛 比赛总结

感觉最近任务还是挺重的，既要参加百度之星的比赛做题改题，又要做CDQ分治的练习题，还要参加学校的NOIP模拟赛写总结。效率还是要更高才好（话说BC的比赛我好像已经很久没动过了QAQ）。

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T1：度度熊保护村庄（hdu6080）

题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6080

题目分析：这题的思路很巧妙啊。一开始以为是求凸包或半平面交之类的，由于没做过多少题计算几何就直接跳过了。赛后看了看网上的大神们的总结，居然写个叉积再用floyd找最小环就可以了！

我们先枚举任意两个度度熊士兵i,j(i!=j)，然后用叉积判断所有房屋是否都在i->j这条有向线段的左侧，如果是的话点i向点j连一条长度为1的边。如果所有点都在线段上则i与j之间连一条双向边。跑一边floyd求最小环，这样最小环上的点就是我们要保留的点，用m-最小环上的点数即可，如果没有环则输出ToT。这题主要是将原图要求的凸包通过合理的建边（确保其凸性）转化成了新图中的最小环，然后用floyd算法解决。

CODE：

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<stdio.h>
#include<algorithm>
using namespace std;

const int maxn=505;
const int oo=1000000001;

struct data
{
int X,Y;
} ;
data p[maxn];
data q[maxn];

int dis[maxn][maxn];
int n,m;

void Add_edge()
{
for (int i=1; i<m; i++)
for (int j=i+1; j<=m; j++)
{
bool L=true,R=true;
for (int k=1; k<=n; k++)
{
int u1=q[i].X-p[k].X;
int v1=q[i].Y-p[k].Y;
int u2=q[j].X-p[k].X;
int v2=q[j].Y-p[k].Y;
int temp=u1*v2-u2*v1;
if (temp>0) L=false;
if (temp<0) R=false;
}
if (L) dis[i][j]=1;
if (R) dis[j][i]=1;
}
}

int Min(int x,int y)
{
return ((x<y)? x:y);
}

void Floyd()
{
for (int k=1; k<=m; k++)
for (int i=1; i<=m; i++) if (dis[i][k]<oo)
for (int j=1; j<=m; j++)
dis[i][j]=Min(dis[i][k]+dis[k][j],dis[i][j]);
}

int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);

scanf("%d",&n);
while (n)
{
for (int i=1; i<=n; i++) scanf("%d%d",&p[i].X,&p[i].Y);
scanf("%d",&m);
for (int i=1; i<=m; i++) scanf("%d%d",&q[i].X,&q[i].Y);
for (int i=1; i<=m; i++)
for (int j=1; j<=m; j++)
dis[i][j]=oo;
Add_edge();
Floyd();
int ans=oo;
for (int i=1; i<=m; i++) ans=Min(ans,dis[i][i]);
if (ans==oo) printf("ToT\n");
else printf("%d\n",m-ans);
n=0;
scanf("%d",&n);
}

return 0;
}

T2：度度熊的王国战略（hdu6081）

题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6081

题目分析：这题很明显就是个全局最小割裸题，用堆优化的全局最小割是O(nmlog(n))的，20s的话勉勉强强吧。不过比赛的时候我看到Clarification那里有人说数据很水，直接用并查集判连通，如果连通的话就割断连向某一个点的所有边，取个min即可；还有人说题意本来就是要割断连向某个点的所有边取min，只是题面有误……然后我就不做这题了。赛后为了凑个AK写了写这题的代码。有一个坑点就是加边的时候要特判自环。

CODE：

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<stdio.h>
#include<algorithm>
using namespace std;

const int maxn=3010;

int fa[maxn];
int val[maxn];

int n,m;

void Up(int x)
{
if ( fa[x]==x ) return;
Up(fa[x]);
fa[x]=fa[ fa[x] ];
}

void Add(int x,int y)
{
Up(x);
Up(y);
if ( fa[x]==fa[y] ) return;
fa[ fa[x] ]=fa[y];
}

int main()
{
freopen("2.in","r",stdin);
freopen("2.out","w",stdout);

scanf("%d%d",&n,&m);
while ( n || m )
{
for (int i=1; i<=n; i++) fa[i]=i,val[i]=0;
for (int i=1; i<=m; i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
if (u==v) continue;
Add(u,v);
val[u]+=w;
val[v]+=w;
}
int x=0;
for (int i=1; i<=n; i++)
{
Up(i);
if (fa[i]==i) x++;
}
if (x>1) printf("0\n");
else
{
int ans=val[1];
for (int i=2; i<=n; i++) ans=min(ans,val[i]);
printf("%d\n",ans);
}
n=m=0;
scanf("%d%d",&n,&m);
}

return 0;
}

T3：度度熊与邪恶大魔王（hdu6082）

题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6082

题目分析：一个很简单的背包，考试时最多人切的就是这题。如果没有b[i]的防御力这个限制，我们可以直接写一个无限背包，用f[i]表示打掉血量为i的怪兽所需要的最少晶石。然后由于b[i]顶多只有10，我们做11次无限背包，用f[i][j]表示打掉血量为i，防御力为j的怪兽所需要的最少晶石，将每一个技能的伤害-j，然后直接做就可以了。如果所有技能中伤害最高的那个小于等于所有怪兽中的最高防御力，则打不死所有怪兽，输出-1。

CODE：

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<stdio.h>
#include<algorithm>
using namespace std;

const int maxb=12;
const int maxm=1010;
const int maxn=100100;
const int oo=1000000001;

int f[maxb][maxm];

int a[maxn];
int b[maxn];

int k[maxm];
int p[maxm];

int n,m;

int main()
{
freopen("3.in","r",stdin);
freopen("3.out","w",stdout);

scanf("%d%d",&n,&m);
while ( n && m )
{
int ma=0,mb=0;
for (int i=1
19f08
; i<=n; i++)
scanf("%d%d",&a[i],&b[i]),ma=max(a[i],ma),mb=max(b[i],mb);
int mp=0;
for (int i=1; i<=m; i++)
scanf("%d%d",&k[i],&p[i]),mp=max(p[i],mp);

if (mp<=mb) printf("-1\n");
else
{
for (int i=0; i<=mb; i++)
{
f[i][0]=0;
for (int j=1; j<=ma; j++) f[i][j]=oo;

for (int j=1; j<=m; j++) if (p[j]>0)
for (int w=1; w<=ma; w++)
if (w<p[j]) f[i][w]=min(f[i][w],k[j]);
else f[i][w]=min(f[i][w],f[i][ w-p[j] ]+k[j]);

for (int j=1; j<=m; j++) p[j]--;
}

long long ans=0;
for (int i=1; i<=n; i++)
ans+=(long long)f[ b[i] ][ a[i] ];
printf("%I64d\n",ans);
}

n=0,m=0;
scanf("%d%d",&n,&m);
}

return 0;
}

T4：度度熊的午饭时光（hdu6083）

题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6083

题目分析：我们发现，如果没有序号和最小和字典序最小这两个限制，这道题就是个01背包。那么序号和怎么搞呢？由于菜品的得分和是严格优先于序号和的（也就是说我们要优先考虑得分和大的方案），所以我们不妨在dp数组里记录一下序号和，然后价值相同的时候用最小的序号和去更新。那么字典序最小怎么办呢？普通的递归打印无法解决字典序问题，我们不妨就在dp数组中记一个长度为100的bool数组，代表每个物品选或不选。序号和相同的时候我们就暴力查看bool数组，但这样会超时。我们发现每一个物品只有选或不选两种状态，开bool数组太浪费时间空间了，我们可以开两个long long类型，每个long long类型压50位，编号小的压在高位，每一位的0代表不选，1代表选。可以证明，这样比较的时候long long值比较大的那个字典序一定更小。

CODE：

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<stdio.h>
#include<algorithm>
using namespace std;

const int maxb=1010;
const int maxn=105;
typedef long long LL;

struct data
{
int val,num;
LL id1,id2;
} f[maxb];

int s[maxn];
int c[maxn];

int b,n;

void Add(int x,int y)
{
if (y<=50) f[x].id1^=(long long)(1LL<<(long long)(50-y));
else f[x].id2^=(long long)(1LL<<(long long)(100-y));
}

bool Judge(int x,int y,int z)
{
LL v1=f[x].id1,v2=f[x].id2;
if (z<=50) v1^=(long long)(1LL<<(long long)(50-z));
else v2^=(long long)(1LL<<(long long)(100-z));
return ( v1>f[y].id1 || ( v1==f[y].id1 && v2>f[y].id2 ) );
}

int Calc()
{
int v=0;
for (int i=49; i>=0; i--) if ( f[b].id1&(1LL<<(long long)i) ) v+=c[50-i];
for (int i=49; i>=0; i--) if ( f[b].id2&(1LL<<(long long)i) ) v+=c[100-i];
return v;
}

void Print()
{
LL v1=f[b].id1,v2=f[b].id2;
int cnt=0;
for (int i=49; i>=0; i--) if ( v1&(1LL<<(long long)i) ) cnt++;
for (int i=49; i>=0; i--) if ( v2&(1LL<<(long long)i) ) cnt++;
if (!cnt) return;
for (int i=49; i>=0; i--)
if ( v1&(1LL<<(long long)i) )
{
cnt--;
printf("%d",50-i);
if (cnt) printf(" ");
}
for (int i=49; i>=0; i--)
if ( v2&(1LL<<(long long)i) )
{
cnt--;
printf("%d",100-i);
if (cnt) printf(" ");
}
printf("\n");
}

int main()
{
freopen("4.in","r",stdin);
freopen("4.out","w",stdout);

scanf("%d",&n);
int cas=0;
scanf("%d%d",&b,&n);
while ( b!=-1 && n!=-1 )
{
for (int i=1; i<=n; i++) scanf("%d%d",&s[i],&c[i]);
for (int i=0; i<=b; i++) f[i].val=f[i].num=f[i].id1=f[i].id2=0;
for (int i=1; i<=n; i++)
for (int j=b; j>=c[i]; j--)
{
if ( f[ j-c[i] ].val+s[i]>f[j].val )
f[j]=f[ j-c[i] ],f[j].val+=s[i],f[j].num+=i,Add(j,i);
else
if ( f[ j-c[i] ].val+s[i]==f[j].val )
if ( f[ j-c[i] ].num+i<f[j].num )
f[j]=f[ j-c[i] ],f[j].val+=s[i],f[j].num+=i,Add(j,i);
else
if ( f[ j-c[i] ].num+i==f[j].num )
if ( Judge(j-c[i],j,i) )
f[j]=f[ j-c[i] ],f[j].val+=s[i],f[j].num+=i,Add(j,i);
}
cas++;
printf("Case #%d:\n",cas);
for (int i=b-1; i>=0; i--)
if ( f[i].val>f[b].val ) b=i;
else
if ( f[i].val==f[b].val )
if ( f[i].num<f[b].num ) b=i;
else
if ( f[i].num==f[b].num )
if ( f[i].id1>f[b].id1 ) b=i;
else
if ( f[i].id1==f[b].id1 )
if ( f[i].id2>f[b].id2 ) b=i;
printf("%d %d\n",f[b].val, Calc() );
Print();
b=n=-1;
scanf("%d%d",&b,&n);
}

return 0;
}

T5：寻找母串（hdu6084）

题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6084

题目分析：这是一道结论题……

先看看长度为n的01偏串个数怎么求，由bzoj1856字符串这题我们可以知道，其个数为Cn2n−Cn2−1n（这就是个很简单的卡特兰数的同构）。然后设给定的01偏串的长度为slen，可以证明一个包含有S的长度为n的01偏串，取出S之后还是一个01偏串，而一个长度为n-slen的01偏串，我们可以将S插入n-slen+1个位置。这样我们算出长度为n-slen的01偏串有多少个，乘以n-slen+1即可。

（PS：由于本蒟蒻还不会log(n)求阶乘，所以我是用打表算阶乘的……）

CODE：

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<stdio.h>
#include<algorithm>
using namespace std;

const int maxn=1000100;
const int maxm=1001;
const long long M=1000000007;
const int f[maxm]={1,641102369,578095319,5832229,259081142,974067448,316220877,690120224,251368199,980250487,682498929,134623568,95936601,933097914,167332441,598816162,336060741,248744620,626497524,288843364,491101308,245341950,565768255,246899319,968999,586350670,638587686,881746146,19426633,850500036,76479948,268124147,842267748,886294336,485348706,463847391,544075857,898187927,798967520,82926604,723816384,156530778,721996174,299085602,323604647,172827403,398699886,530389102,294587621,813805606,67347853,497478507,196447201,722054885,228338256,407719831,762479457,746536789,811667359,778773518,27368307,438371670,59469516,5974669,766196482,606322308,86609485,889750731,340941507,371263376,625544428,788878910,808412394,996952918,585237443,1669644,361786913,480748381,595143852,837229828,199888908,526807168,579691190,145404005,459188207,534491822,439729802,840398449,899297830,235861787,888050723,656116726,736550105,440902696,85990869,884343068,56305184,973478770,168891766,804805577,927880474,876297919,934814019,676405347,567277637,112249297,44930135,39417871,47401357,108819476,281863274,60168088,692636218,432775082,14235602,770511792,400295761,697066277,421835306,220108638,661224977,261799937,168203998,802214249,544064410,935080803,583967898,211768084,751231582,972424306,623534362,335160196,243276029,554749550,60050552,797848181,395891998,172428290,159554990,887420150,970055531,250388809,487998999,856259313,82104855,232253360,513365505,244109365,1559745,695345956,261384175,849009131,323214113,747664143,444090941,659224434,80729842,570033864,664989237,827348878,195888993,576798521,457882808,731551699,212938473,509096183,827544702,678320208,677711203,289752035,66404266,555972231,195290384,97136305,349551356,785113347,83489485,66247239,52167191,307390891,547665832,143066173,350016754,917404120,296269301,996122673,23015220,602139210,748566338,187348575,109838563,574053420,105574531,304173654,542432219,34538816,325636655,437843114,630621321,26853683,933245637,616368450,238971581,511371690,557301633,911398531,848952161,958992544,925152039,914456118,724691727,636817583,238087006,946237212,910291942,114985663,492237273,450387329,834860913,763017204,368925948,475812562,740594930,45060610,806047532,464456846,172115341,75307702,116261993,562519302,268838846,173784895,243624360,61570384,481661251,938269070,95182730,91068149,115435332,495022305,136026497,506496856,710729672,113570024,366384665,564758715,270239666,277118392,79874094,702807165,112390913,730341625,103056890,677948390,339464594,167240465,108312174,839079953,479334442,271788964,135498044,277717575,591048681,811637561,353339603,889410460,839849206,192345193,736265527,316439118,217544623,788132977,618898635,183011467,380858207,996097969,898554793,335353644,54062950,611251733,419363534,965429853,160398980,151319402,990918946,607730875,450718279,173539388,648991369,970937898,500780548,780122909,39052406,276894233,460373282,651081062,461415770,358700839,643638805,560006119,668123525,686692315,673464765,957633609,199866123,563432246,841799766,385330357,504962686,954061253,128487469,685707545,299172297,717975101,577786541,318951960,773206631,306832604,204355779,573592106,30977140,450398100,363172638,258379324,472935553,93940075,587220627,776264326,793270300,291733496,522049725,579995261,335416359,142946099,472012302,559947225,332139472,499377092,464599136,164752359,309058615,86117128,580204973,563781682,954840109,624577416,895609896,888287558,836813268,926036911,386027524,184419613,724205533,403351886,715247054,716986954,830567832,383388563,68409439,6734065,189239124,68322490,943653305,405755338,811056092,179518046,825132993,343807435,985084650,868553027,148528617,160684257,882148737,591915968,701445829,529726489,302177126,974886682,241107368,798830099,940567523,11633075,325334066,346091869,115312728,473718967,218129285,878471898,180002392,699739374,917084264,856859395,435327356,808651347,421623838,105419548,59883031,322487421,79716267,715317963,429277690,398078032,316486674,384843585,940338439,937409008,940524812,947549662,833550543,593524514,996164327,987314628,697611981,636177449,274192146,418537348,925347821,952831975,893732627,1277567,358655417,141866945,581830879,987597705,347046911,775305697,125354499,951540811,247662371,343043237,568392357,997474832,209244402,380480118,149586983,392838702,309134554,990779998,263053337,325362513,780072518,551028176,990826116,989944961,155569943,596737944,711553356,268844715,451373308,379404150,462639908,961812918,654611901,382776490,41815820,843321396,675258797,845583555,934281721,741114145,275105629,666247477,325912072,526131620,252551589,432030917,554917439,818036959,754363835,795190182,909210595,278704903,719566487,628514947,424989675,321685608,50590510,832069712,198768464,702004730,99199382,707469729,747407118,302020341,497196934,5003231,726997875,382617671,296229203,183888367,703397904,552133875,732868367,350095207,26031303,863250534,216665960,561745549,352946234,784139777,733333339,503105966,459878625,803187381,16634739,180898306,68718097,985594252,404206040,749724532,97830135,611751357,31131935,662741752,864326453,864869025,167831173,559214642,718498895,91352335,608823837,473379392,385388084,152267158,681756977,46819124,313132653,56547945,442795120,796616594,256141983,152028387,636578562,385377759,553033642,491415383,919273670,996049638,326686486,160150665,141827977,540818053,693305776,593938674,186576440,688809790,565456578,749296077,519397500,551096742,696628828,775025061,370732451,164246193,915265013,457469634,923043932,912368644,777901604,464118005,637939935,956856710,490676632,453019482,462528877,502297454,798895521,100498586,699767918,849974789,811575797,438952959,606870929,907720182,179111720,48053248,508038818,811944661,752550134,401382061,848924691,764368449,34629406,529840945,435904287,26011548,208184231,446477394,206330671,366033520,131772368,185646898,648711554,472759660,523696723,271198437,25058942,859369491,817928963,330711333,724464507,437605233,701453022,626663115,281230685,510650790,596949867,295726547,303076380,465070856,272814771,538771609,48824684,951279549,939889684,564188856,48527183,201307702,484458461,861754542,326159309,181594759,668422905,286273596,965656187,44135644,359960756,936229527,407934361,267193060,456152084,459116722,124804049,262322489,920251227,816929577,483924582,151834896,167087470,490222511,903466878,361583925,368114731,339383292,388728584,218107212,249153339,909458706,322908524,202649964,92255682,573074791,15570863,94331513,744158074,196345098,334326205,9416035,98349682,882121662,769795511,231988936,888146074,137603545,582627184,407518072,919419361,909433461,986708498,310317874,373745190,263645931,256853930,876379959,702823274,147050765,308186532,175504139,180350107,797736554,606241871,384547635,273712630,586444655,682189174,666493603,946867127,819114541,502371023,261970285,825871994,126925175,701506133,314738056,341779962,561011609,815463367,46765164,49187570,188054995,957939114,64814326,933376898,329837066,338121343,765215899,869630152,978119194,632627667,975266085,435887178,282092463,129621197,758245605,827722926,201339230,918513230,322096036,547838438,985546115,852304035,593090119,689189630,555842733,567033437,469928208,212842957,117842065,404149413,155133422,663307737,208761293,206282795,717946122,488906585,414236650,280700600,962670136,534279149,214569244,375297772,811053196,922377372,289594327,219932130,211487466,701050258,398782410,863002719,27236531,217598709,375472836,810551911,178598958,247844667,676526196,812283640,863066876,857241854,113917835,624148346,726089763,564827277,826300950,478982047,439411911,454039189,633292726,48562889,802100365,671734977,945204804,508831870,398781902,897162044,644050694,892168027,828883117,277714559,713448377,624500515,590098114,808691930,514359662,895205045,715264908,628829100,484492064,919717789,513196123,748510389,403652653,574455974,77123823,172096141,819801784,581418893,15655126,15391652,875641535,203191898,264582598,880691101,907800444,986598821,340030191,264688936,369832433,785804644,842065079,423951674,663560047,696623384,496709826,161960209,331910086,541120825,951524114,841656666,162683802,629786193,190395535,269571439,832671304,76770272,341080135,421943723,494210290,751040886,317076664,672850561,72482816,493689107,135625240,100228913,684748812,639655136,906233141,929893103,277813439,814362881,562608724,406024012,885537778,10065330,60625018,983737173,60517502,551060742,804930491,823845496,727416538,946421040,678171399,842203531,175638827,894247956,538609927,885362182,946464959,116667533,749816133,241427979,871117927,281804989,163928347,563796647,640266394,774625892,59342705,256473217,674115061,918860977,322633051,753513874,393556719,304644842,767372800,161362528,754787150,627655552,677395736,799289297,846650652,816701166,687265514,787113234,358757251,701220427,607715125,245795606,600624983,10475577,728620948,759404319,36292292,491466901,22556579,114495791,647630109,586445753,482254337,718623833,763514207,66547751,953634340,351472920,308474522,494166907,634359666,172114298,865440961,364380585,921648059,965683742,260466949,117483873,962540888,237120480,620531822,193781724,213092254,107141741,602742426,793307102,756154604,236455213,362928234,14162538,753042874,778983779,25977209,49389215,698308420,859637374,49031023,713258160,737331920,923333660,804861409,83868974,682873215,217298111,883278906,176966527,954913,105359006,390019735,10430738,706334445,315103615,567473423,708233401,48160594,946149627,346966053,281329488,462880311,31503476,185438078,965785236,992656683,916291845,881482632,899946391,321900901,512634493,303338827,121000338,967284733,492741665,152233223,165393390,680128316,917041303,532702135,741626808,496442755,536841269,131384366,377329025,301196854,859917803,676511002,373451745,847645126,823495900,576368335,73146164,954958912,847549272,241289571,646654592,216046746,205951465,3258987,780882948,822439091,598245292,869544707,698611116};
typedef long long LL;

char s[maxn];
int t,n,slen;
LL X,Y;

LL Get(int x)
{
int y=x/1000000;
LL val=f[y];
y*=1000000;
for (int i=y+1; i<=x; i++) val=val*(long long)i%M;
return val;
}

void Exgcd(LL a,LL b)
{
if (!b) X=1,Y=0;
else
{
Exgcd(b,a%b);
LL c=a/b;
LL u=Y,v=X-c*Y;
X=u,Y=v;
}
}

int main()
{
freopen("5.in","r",stdin);
freopen("5.out","w",stdout);

scanf("%d",&t);
while (t--)
{
scanf("%d %s\n",&n,&s);
int slen=strlen(s);
if ( n<slen || n&1 ) printf("0\n");
else
{
n-=slen;
if (!n)
{
printf("1\n");
continue;
}

LL val=Get(n);
LL dec=val;

LL temp=Get(n>>1);
Exgcd(temp,M);
val=val*X%M*X%M;

temp=Get((n>>1)-1);
Exgcd(temp,M);
dec=dec*X%M;

temp=Get((n>>1)+1);
Exgcd(temp,M);
dec=dec*X%M;

LL ans=(val-dec+M)%M;
ans=ans*(long long)(n+1)%M;
printf("%I64d\n",ans);
}
}

return 0;
}

总结：这次的比赛主要是晋级初赛用的，做对一题就可以了，而且时间有36个小时，所以慢悠悠地做题根本不虚。比赛的时候做了3,4,5题，第二题觉得太狗血就没做，T1想不出来。感觉这套题比我们学校的辣鸡NOIP模拟赛的题目质量要高很多，主要是考察基础知识，既没有什么高深算法，思维难度也适中。

希望接下来的比赛能正常发挥吧（起码要到纪念衫呵呵）。