http://poj.org/problem?id=2455 (题目链接)
题意
给出一张n个点，p条边的无向图，需要从1号节点走到n号节点一共T次，每条边只能经过1次，问T次经过的最大的边最小是多少。
Solution
很显然，二分答案，然后建图跑最大流即可。
细节
双向边？
代码
// poj2455
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define LL long long
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;const int maxn=300,maxm=40010;
struct edge {int to,next,w;}e[maxm<<1];
int head[maxn],d[maxn];
int cnt,n,p,T,u[maxm],v[maxm],w[maxm],ans;void link(int u,int v,int w) {e[++cnt]=(edge){v,head[u],w};head[u]=cnt;e[++cnt]=(edge){u,head[v],w};head[v]=cnt;
}
void build(int k) {cnt=1;memset(head,0,sizeof(head));for (int i=1;i<=p;i++)if (w[i]<=k) link(u[i],v[i],1);
}
bool bfs() {memset(d,-1,sizeof(d));queue<int> q;q.push(1);d[1]=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[n]>0;
}
int dfs(int x,int f) {if (x==n || f==0) return f;int used=0,w;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;
}
void Dinic() {while (bfs()) ans+=dfs(1,inf);
}
int main() {scanf("%d%d%d",&n,&p,&T);int L=inf,R=0;for (int i=1;i<=p;i++)scanf("%d%d%d",&u[i],&v[i],&w[i]),L=min(L,w[i]),R=max(R,w[i]);int res=0;while (L<=R) {int mid=(L+R)>>1;build(mid);ans=0;Dinic();if (ans>=T) res=mid,R=mid-1;else L=mid+1;}printf("%d",res);return 0;
}
