标签:int push head out bre while div oid ace
http://www.lydsy.com/JudgeOnline/problem.php?id=1570 (题目链接)
给出$m$个航班,每天只能做一次飞机,有$T$人从起点到终点,问最晚到达的人最早什么时候到。
枚举答案分层建图最大流判断即可。之前的流量不要清空。
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// bzoj1570 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #include<cmath> #include<queue> #define LL long long #define inf (1ll<<30) #define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout) using namespace std; const int maxn=1000010; int head[maxn],n,m,T,et,ans,cnt=1; struct data {int u,v,w;}t[maxn]; struct edge {int to,next,w;}e[maxn]; void link(int u,int v,int w) { e[++cnt]=(edge){v,head[u],w};head[u]=cnt; e[++cnt]=(edge){u,head[v],0};head[v]=cnt; } namespace Dinic { int d[maxn],s,t; bool bfs(int s,int t) { memset(d,-1,sizeof(d)); queue<int> q;q.push(s);d[s]=0; while (!q.empty()) { int x=q.front();q.pop(); for (int i=head[x];i;i=e[i].next) if (e[i].w && d[e[i].to]<0) d[e[i].to]=d[x]+1,q.push(e[i].to); } return d[t]>0; } int dfs(int x,int f) { if (x==t || f==0) return f; int w,used=0; for (int i=head[x];i;i=e[i].next) if (e[i].w && d[e[i].to]==d[x]+1) { w=dfs(e[i].to,min(e[i].w,f-used)); used+=w;e[i].w-=w;e[i^1].w+=w; if (used==f) return used; } if (!used) d[x]=-1; return used; } int main(int x,int y) { s=x,t=y;int flow=0; while (bfs(x,y)) flow+=dfs(x,inf); return flow; } } int main() { scanf("%d%d%d",&n,&m,&T); for (int i=1;i<=m;i++) scanf("%d%d%d",&t[i].u,&t[i].v,&t[i].w); link(0,1,T);et=1000000; for (int i=0;1;i++) { for (int j=1;j<=n;j++) link(i*n+j,(i+1)*n+j,inf); for (int j=1;j<=m;j++) link(i*n+t[j].u,(i+1)*n+t[j].v,t[j].w); link((i+1)*n,et,inf); ans+=Dinic::main(0,et); if (ans==T) {printf("%d",i);break;} } return 0; }
【bzoj1507】 JSOI2008—Blue Mary的旅行
标签:int push head out bre while div oid ace
原文地址:http://www.cnblogs.com/MashiroSky/p/6616682.html