poj 2455

说实话我都不知道为什么要写这题题解。毕竟代码都不是我写的，
嗯
思路很好想 二分+网络流判定可行性，这个想到了。

然后决定学习dinic

所以当个备忘录，明天求教

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <queue>  
using namespace std;  

const int N = 210;  
const int INF = 0x3f3f3f3f;  
struct edge  
{  
    int to, cap, next;  
}g[N*1000];  
int head[N], iter[N], level[N], d[N*400][3];  
int cnt;  
void add_edge(int v, int u, int cap)  
{  
    g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;  
    g[cnt].to = v, g[cnt].cap = cap, g[cnt].next = head[u], head[u] = cnt++; //无向图时反向边容量不为0  
}  
bool bfs(int s, int t)  
{  
    memset(level, -1, sizeof level);  
    queue<int> que;  
    level[s] = 0;  
    que.push(s);  
    while(! que.empty())  
    {  
        int v = que.front(); que.pop();  
        for(int i = head[v]; i != -1; i = g[i].next)  
        {  
            int u = g[i].to;  
            if(g[i].cap > 0 && level[u] < 0)  
            {  
                level[u] = level[v] + 1;  
                que.push(u);  
            }  
        }  
    }  
    return level[t] == -1;  
}  
int dfs(int v, int t, int f)  
{  
    if(v == t) return f;  
    for(int &i = iter[v]; i != -1; i = g[i].next)  
    {  
        int u = g[i].to;  
        if(g[i].cap > 0 && level[v] < level[u])  
        {  
            int d = dfs(u, t, min(f, g[i].cap));  
            if(d > 0)  
            {  
                g[i].cap -= d, g[i^1].cap += d;  
                return d;  
            }  
        }  
    }  
    return 0;  
}  
int dinic(int s, int t)  
{  
    int flow = 0, f;  
    while(true)  
    {  
        if(bfs(s, t)) return flow;  
        memcpy(iter, head, sizeof head);  
        while(f = dfs(s, t, INF), f > 0)  
            flow += f;  
    }  
}  
int main()  
{  
    int n, p, t;  
    while(~ scanf("%d%d%d", &n, &p, &t))  
    {  
        int maxx = -1, minn = INF;  
        for(int i = 0; i < p; i++)  
        {  
            scanf("%d%d%d", &d[i][0], &d[i][1], &d[i][2]);  
            maxx = max(maxx, d[i][2]);  
            minn = min(minn, d[i][2]);  
        }  
        int l = minn, r = maxx, res;  
        while(l <= r)  
        {  
            int mid = (l + r) >> 1;  
            cnt = 0;  
            memset(head, -1, sizeof head);  
            for(int i = 0; i < p; i++)  
                if(d[i][2] <= mid)  
                    add_edge(d[i][0], d[i][1], 1);  
            if(dinic(1, n) >= t) r = mid - 1, res = mid;  
            else l = mid + 1;  
        }  
        printf("%d\n", res);  
    }  
    return 0;  
}