Codeforces Round #380 (Div. 2, Rated, Based on Technocup 2017 - Elimination Round 2)

链接：http://codeforces.com/contest/738

A题：水题

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=1e9+10;

int n;
char s[maxn];

int main(){
//freopen("in.txt","r",stdin);
while(cin>>n>>(s+1)){
for(int i=1;i<=n;){
int j=i+1;
if(j+1<=n&&s[i]=='o'&&s[j]=='g'&&s[j+1]=='o'){
while(j+1<=n&&s[j]=='g'&&s[j+1]=='o') j+=2;
s[i]=s[i+1]=s[i+2]='*';
for(int k=i+3;k<j;k++) s[k]='#';
}
i=j;
}
//		cout<<(s+1)<<endl;
for(int i=1;i<=n;i++) if(s[i]!='#') printf("%c",s[i]);puts("");
}
return 0;
}

B题：水题

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const int maxn=1100;
const int INF=1e9+10;

int n,m;
int a[maxn][maxn];
int L[maxn],R[maxn],D[maxn],U[maxn];

int main(){
//freopen("in.txt","r",stdin);
while(cin>>n>>m){
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
scanf("%d",&a[i][j]);
for(int i=1;i<=n;i++){
L[i]=m+1;R[i]=0;
for(int j=1;j<=m;j++){
if(!a[i][j]) continue;
L[i]=min(L[i],j);
R[i]=max(R[i],j);
}
}
for(int j=1;j<=m;j++){
U[j]=n+1;D[j]=0;
for(int i=1;i<=n;i++){
if(!a[i][j]) continue;
U[j]=min(U[j],i);
D[j]=max(D[j],i);
}
}
int res=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(!a[i][j]) res+=(L[i]<j)+(R[i]>j)+(U[j]<i)+(D[j]>i);
}
}
cout<<res<<endl;
}
return 0;
}

C题：水题，二分

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=1e9+10;
const double EPS=1e-10;

struct Node{
int c,v;
friend bool operator<(Node A,Node B){
return A.c==B.c?A.v>B.v:A.c<B.c;
}
};Node p[maxn];
int n,k,s,t;
int g[maxn];

ll get(int v,int d){
if(v<d) return -1;
int v2=2*(v-d);
if(v2>d*2) v2=d*2;
int rest=d-v2/2;
int v1=max(0,rest);
return 2LL*v1+v2/2;
}

bool check(int v){
ll tt=0;
for(int i=1;i<=k;i++){
int d=g[i]-g[i-1];
ll tmp=get(v,d);
if(tmp==-1) return 0;
tt+=tmp;
952a
if(tt>t) return 0;
}
return 1;
}

int bin(int l,int r){
int id=INF;
while(l<=r){
int m=(l+r)>>1;
if(check(p[m].v)) id=min(id,m),r=m-1;
else l=m+1;
}
if(id==INF) return -1;
return p[id].c;
}

int main(){
//	freopen("in.txt","r",stdin);
while(cin>>n>>k>>s>>t){
for(int i=1;i<=n;i++) scanf("%d%d",&p[i].c,&p[i].v);
for(int i=1;i<=k;i++) scanf("%d",&g[i]);
g[0]=0;
g[++k]=s;
sort(g,g+k+1);
sort(p+1,p+n+1);
int pn=1;
for(int i=2;i<=n;i++){
if(p[i].v<=p[pn].v) continue;
p[++pn]=p[i];
}
printf("%d\n",bin(1,pn));
}
return 0;
}

D题：水题，贪心

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=1e9+10;

int n,a,b,k;
char s[maxn];
struct Node{
int len,cnt;
int l,r;
friend bool operator<(Node A,Node B){
return A.len<B.len;
}
};Node p[maxn];int pn;
int ans[maxn],ansn;

void solve(int l,int r,int c){
int len=r-l+1;
if(len<2*b-1) ans[++ansn]=l+b-1;
else{
int cnt=len/b;
for(int i=l+b-1;i<=r;i+=b){
ans[++ansn]=i;
cnt--;
if(cnt<c) break;
}
}
}

int main(){
//	freopen("in.txt","r",stdin);
while(cin>>n>>a>>b>>k){
scanf("%s",s+1);
pn=0;
for(int i=1;i<=n;){
if(s[i]=='0'){
int j=i+1;
while(j<=n&&s[j]=='0') j++;
if(j-i>=b) p[++pn].len=j-i,p[pn].l=i,p[pn].r=j-1;
i=j;
}
else i++;
}
sort(p+1,p+pn+1);
int ta=a;
for(int i=1;i<=pn;i++){
int c=p[i].len/b;
if(c<=ta) p[i].cnt=c,ta-=c;
else p[i].cnt=ta,ta=0;
}
ansn=0;
for(int i=pn;i>=1;i--){
solve(p[i].l,p[i].r,p[i].cnt);
if(p[i].cnt) break;
}
printf("%d\n",ansn);
for(int i=1;i<=ansn;i++) printf("%d ",ans[i]);puts("");
}
return 0;
}

E题：水题，给一颗有根树，给定每个结点的祖先结点的数量序列，求使得序列合法的最小改变。首先明确合法的序列是：有且只有一个0，排序之后序列相邻数之差<=1。因此只要把其它的0设为INF，然后排序，从前往后贪心，用后面的来填前面的空位就好了，用链表模拟一下就可以了。

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=1e9+10;

int n,a[maxn],s;
int val[maxn],L[maxn],R[maxn];
int tot,rt;

int newnode(int v){
++tot;
val[tot]=v;
L[tot]=R[tot]=0;
return tot;
}

void link(int l,int r){
R[l]=r;L[r]=l;
}

int main(){
//freopen("in.txt","r",stdin);
while(cin>>n>>s){
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
int res=0;
if(a[s]) a[s]=0,res=1;
sort(a+1,a+n+1);
for(int i=2;i<=n;i++) if(!a[i]) a[i]=INF;
sort(a+1,a+n+1);
tot=0;
rt=newnode(a[1]);
for(int i=2;i<=n;i++){
int p=newnode(a[i]);
link(tot-1,p);
}
int l=rt,r=tot;
while(l!=r){
if(val[R[l]]<=val[l]+1) l=R[l];
else{
int p=r;
r=L[r];R[r]=0;
link(p,R[l]);
link(l,p);
val[p]=val[l]+1;
res++;
if(l==r) break;
l=R[l];
}
}
if(val[r]>val[l]+1) res++;
cout<<res<<endl;
}
return 0;
}

F题：博弈dp，样例没看懂。。留坑待补。。。