2018-2019 ACM-ICPC, Asia Nanjing Regional Contest 一些题解
2018-11-18 17:20
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版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/lingzidong/article/details/84202335
比较顺利的一场……
A Adrien and Austin
博弈水题,规律是N是偶数(包括0)K等于1的时候后手必胜,剩下的情况都是先手胜
J Prime Game
题意概括一下就是求所有区间里的数字的不同素因子的个数的和。
应该是花式算贡献的题,我们记录一下每个素因子出现的最后位置last[k],那么对于ai,如果他存在素因子一个k,那么他首先向后贡献了n - i +1,但是和离他最近的last[k]位置是有重复的,最后的贡献就是(n-i+1)*(i-last[i])
#include <bits/stdc++.h> #include<cstring> #define ll long long using namespace std; const int MAXN = 1e6 + 6; int prime[MAXN+10]; bool isprime[MAXN]; void getPrime() { memset(prime,0,sizeof(prime)); memset(isprime,0,sizeof(isprime)); for(int i = 2;i<=MAXN;i++) { if(!prime[i])isprime[i] = 1, prime[++prime[0]] = i; for(int j = 1;j<=prime[0] && prime[j] <= MAXN/i;j++) { prime[prime[j]*i] = 1; if(i % prime[j] == 0) break; } } } ll a[MAXN]; int sz[MAXN]; ll fac[100][2]; int cnt = 0; int getFac(ll x) { cnt = 0; ll tmp = x; for(int i = 1;prime[i] <= tmp/prime[i];i++) { fac[cnt][1] = 0; if(tmp%prime[i] == 0) { fac[cnt][0] = prime[i]; while(tmp % prime[i] == 0) { fac[cnt][1]++; tmp /= prime[i]; } cnt++; if(isprime[tmp]) break; } } if(tmp != 1) { fac[cnt][0] = tmp; fac[cnt++][1] =1; } return cnt; } map<ll,int> m; int main() { //freopen("input.txt","r",stdin); getPrime(); //cout<<prime[0]<<endl; int n; scanf("%d",&n); ll ans = 0; for(int i = 1;i<=n;i++) { scanf("%lld",&a[i]); getFac(a[i]); for(int j = 0;j<cnt;j++){ ans+=(n-i+1)*(1LL*i-m[fac[j][0]]); m[fac[j][0]] = i; //cout<<fac[j][0]<<' '<<ans<<endl; } } printf("%lld\n",ans); return 0; }
I - Magic Potion
网络流简单二分图模型建模,多了两个限制。
第一个限制,左边的点最多只能流出2
第二个限制,附加的只有k的流量
第一个限制让我们不得不让流进左边的点的流量限制在2,第二个限制我们就不能让所有的附加边连接到超级源点了,需要限流。
我们附加一个源点,超级源点向这个点连接一个流量k的点,这个源点再向左边的点连接容量为1的点。表示只能使用K个,每个人只能使用一次。剩下的就很简单了。
#include <bits/stdc++.h> #include<cstring> #define ll long long using namespace std; const int MAXN = 3010; const int MAXM = 1200001; const int INF = 0x3f3f3f3f; struct Edge { int to,next,cap,flow,from; }edge[MAXM]; int tot,head[MAXN]; void init() { memset(head,-1,sizeof(head)); tot = 2; } void addedge(int u,int v,int w,int rw = 0) { edge[tot].to = v,edge[tot].from = u,edge[tot].cap = w,edge[tot].flow = 0; edge[tot].next = head[u];head[u] = tot++; edge[tot].to = u,edge[tot].from = v,edge[tot].cap = rw,edge[tot].flow = 0; edge[tot].next = head[v];head[v] = tot++; } int Q[MAXN]; int dep[MAXN],cur[MAXN],sta[MAXN]; bool bfs(int s,int t,int n) { int front = 0,tail = 0; 1fff7 memset(dep,-1,sizeof(dep)); dep[s] = 0; Q[tail++] = s; while(front < tail) { int u =Q[front++]; for(int i = head[u];i != -1;i=edge[i].next) { int v = edge[i].to; if(edge[i].cap > edge[i].flow && dep[v] == -1) { dep[v]= dep[u] + 1; if(v == t) return 1; Q[tail++] = v; } } } return 0; } int dinic(int s,int t,int n) { int maxflow = 0; while(bfs(s,t,n)) { for(int i = 0;i<n;i++) cur[i] = head[i]; int u = s,tail = 0; while(cur[s] != -1) { if(u == t) { int tp = INF; for(int i = tail-1;i >= 0;i--) { tp = min(tp,edge[sta[i]].cap -edge[sta[i]].flow); } maxflow += tp; for(int i = tail-1;i >= 0;i--) { edge[sta[i]].flow += tp; edge[sta[i]^1].flow -= tp; if(edge[sta[i]].cap -edge[sta[i]].flow == 0) tail = i; } u = edge[sta[tail]^1].to; } else if(cur[u] != -1 && edge[cur[u]].cap > edge[cur[u]].flow && dep[u]+1 == dep[edge[cur[u]].to]) { sta[tail++] = cur[u]; u = edge[cur[u]].to; } else { while(u != s && cur[u] == -1) { u = edge[sta[--tail]^1].to; } cur[u] = edge[cur[u]].next; } } } return maxflow; } void debug() { for(int i = 0;i<tot;i += 2) { printf("%d->%d : cap = %d flow = %d\n",edge[i].from,edge[i].to,edge[i].cap,edge[i].flow); } } int main() { int n,m,k; cin>>n>>m>>k; init(); int s = 0,s1 = (n + m)+1,t = (n +m)+2; addedge(s,s1,k); for(int i = 1;i<=n;i++) { addedge(s,i,1); addedge(s1,i,1); int t; cin>>t; for(int j = 0;j<t;j++) { int x; cin>>x; addedge(i,x+n,1); } } for(int i = 1;i<=m;i++) addedge(i+n,t,1); int ans = dinic(s,t,(n+m)+4); //debug(); cout<<ans<<endl; return 0; }
G Pyramid
无敌的队友怒推规律,发现是(n4)n \choose 4(4n),剩下的就是输出了。
D Country Meow
裸最小球覆盖……怒抄模板就过了
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