poj2455Secret Milking Machine(二分 ＋ maxflow)

求出T条从1到n的路径，要求最长的单条边的长度最短，路径不能重复。

第一感觉就是二分最大路径上限，然后建图，满足条件的边容量设为1，求最大流。

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2016
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
typedef pair<ii,int> iii;
const double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 210;
const int maxm = 160000 + 10;
struct Edge {
    int from, to, cap, flow, nxt;
    Edge(){}
    Edge(int from,int to,int cap, int flow, int nxt) : 
        from(from),to(to),cap(cap),flow(flow),nxt(nxt) {}
}edges[maxm];
struct Item {
    int u, v, cost;
}item[maxm];
int dis[maxn];
int head[maxn], ecnt;//链式前向星
int s, t;//源点，汇点
int n, m, T;//点，边
bool spfa(int s,int t) {
    memset(dis, -1,sizeof dis);
    dis[s] = 0;
    queue<int> que;
    que.push(s);
    while(!que.empty()) {
        int u = que.front();
        que.pop();
        for (int i = head[u];~i;i = edges[i].nxt) {
            Edge& e = edges[i];
            if (e.cap > e.flow && dis[e.to] == -1) {
                dis[e.to] = dis[u] + 1;
                que.push(e.to);
            }
        }
    }
    return dis[t] != -1;
}
int dfs(int u,int a) {
    if (u == t || a == 0) return a;
    int flow = 0, f;
    for (int i = head[u];~i;i = edges[i].nxt) {
        Edge& e = edges[i];
        if (dis[e.to] == dis[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
            flow += f;
            a -= f;
            e.flow += f;
            edges[i^1].flow -= f;
            if (a <= 0) break;
        }
    }
    if (flow == 0) dis[u] = 0;
    return flow;
}
int dinic(int s,int t) {
    int ret = 0;
    while(spfa(s, t)) {
        ret += dfs(s, INF);
    }
    return ret;
}
void Getmap(int mid) {
    ecnt = 0;
    memset(head, -1,sizeof head);

    for (int i = 1;i <= m;++i) {
        if (item[i].cost <= mid) {
            edges[ecnt] = Edge(item[i].u, item[i].v, 1, 0, head[item[i].u]), head[item[i].u] = ecnt++;
            edges[ecnt] = Edge(item[i].v, item[i].u, 0, 0, head[item[i].v]), head[item[i].v] = ecnt++;
           
            edges[ecnt] = Edge(item[i].u, item[i].v, 0, 0, head[item[i].u]), head[item[i].u] = ecnt++;
            edges[ecnt] = Edge(item[i].v, item[i].u, 1, 0, head[item[i].v]), head[item[i].v] = ecnt++;
        }
    }
}
int main()
{    
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    while(~scanf("%d%d%d",&n,&m,&T)) {
        s = 1, t = n;
        int high = 0, low = inf;
        for (int i = 1;i <= m;++i) {
            scanf("%d%d%d",&item[i].u, &item[i].v,&item[i].cost);
            high = max(high, item[i].cost);
            low = min(low, item[i].cost);
        }
        high++;
        int tmp;
        while(low <= high) {
            int mid = (low + high) >> 1;
            Getmap(mid);
            int ans = dinic(s, t);
            if (ans >= T) {
                high = mid - 1;
                tmp = mid;
            }
            else low = mid + 1;
        }
        printf("%d\n", tmp);
        // printf("%d\n", low);
        // printf("%d\n", high);
    }
    return 0;
}