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题目链接:
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> #include <map> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar()); for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + ‘0‘); putchar(‘\n‘); } const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=1e5+10; const int maxn=210; const double eps=1e-12; int n,m,l,num[maxn],a[maxn][maxn],v[maxn],p[maxn]; int cap[2*maxn][2*maxn],path[2*maxn],minflow[2*maxn],s,e; queue<int>qu; struct node { int va,t,id; }po[210]; int cmp(node x,node y) { return x.t<y.t; } int bfs() { mst(path,-1); qu.push(s); path[s]=0;minflow[s]=inf; while(!qu.empty()) { int fr=qu.front(); qu.pop(); for(int i=1;i<=e;i++) { if(path[i]==-1&&cap[fr][i]) { minflow[i]=min(minflow[fr],cap[fr][i]); path[i]=fr; qu.push(i); } } } if(path[e]==-1)return 0; return minflow[e]; } int maxflow(int x) { mst(cap,0); int sum=0; For(i,1,m) { if(po[a[i][num[i]]].t<=po[x].t) { sum+=v[i]; cap[s][i]=v[i]; For(j,1,num[i])cap[i][m+a[i][j]]=inf; } } for(int i=1;i<=x;i++) { cap[i+m][e]=po[i].va; } int ans=0; while(bfs()) { int temp=minflow[e]; int cur=e; while(cur!=s) { int fa=path[cur]; cap[fa][cur]-=temp; cap[cur][fa]+=temp; cur=fa; } ans+=temp; } //cout<<sum<<" "<<ans<<" &&&&"<<endl; if(sum-ans>=l) { printf("%d %d\n",po[x].t,sum-ans); return 1; } return 0; } int solve() { s=0;e=n+m+1; int flag=1; For(i,1,n) { if(i!=n&&po[i].t==po[i+1].t)continue; else { if(maxflow(i)){flag=0;break;} } } if(flag)printf("impossible\n"); } int main() { int t,Case=0; read(t); while(t--) { printf("Case #%d: ",++Case); read(n);read(m);read(l); For(i,1,n) { read(po[i].va);read(po[i].t); po[i].id=i; } sort(po+1,po+n+1,cmp); For(i,1,n)p[po[i].id]=i; For(i,1,m) { read(v[i]);read(num[i]); For(j,1,num[i]) { read(a[i][j]); a[i][j]=p[a[i][j]]; } sort(a[i]+1,a[i]+num[i]+1); } solve(); } return 0; }
hdu-5855 Less Time, More profit(最大权闭合子图)
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5777786.html