poj 1797 Heavy Transportation

                                                                      Heavy Transportation

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 44206		Accepted: 11585

Description

Background
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know.

Problem
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo's place) to crossing n (the customer's place). You may assume that there is at least one path. All streets can be travelled in both directions.

Input

The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.

Sample Input

1
3 3
1 2 3
1 3 4
2 3 5

Sample Output

Scenario #1:
4

Source

//刚做了一个最大值的最小值，这题又是最小值的最大值.
#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int maxn = 1010;
int G[maxn][maxn];  //其中G[i][j]表示十字路口i和十字路口j之间的街道最大承载量值;
int dis[maxn];   //dis[i]表示起点1到点i的所有路径中的最大承载量;
int visit[maxn];  //visit[i]表示点i是否被访问过;

void dijkstra(int n){
    int k;
    for(int i = 1; i <= n; i++){  //初始时刻,如果点i与1点直接相连,则最大承载量为这条路的承载量,否则为最小值0;
        dis[i] = G[1][i];
    }
    dis[1] = 0; //自己到自己的承载量没有意义，这一步可省略
    memset(visit, 0, sizeof(visit));
    while(true){
        k = 0;
        for(int i = 1; i <= n; i++){  //从未被访问的集合中找出一个点,它是当前所有点为确定dis中最大的
            if(!visit[i] && (k == 0 || dis[k]<dis[i]))
                k = i;
        }
        if(k == 0)  //所有点都被访问过，跳出循环;
            break;
        visit[k] = 1;  //这个点设置为访问，表明已经缺点了点k到1点的所有路径最大承载量值;
        for(int i = 1; i <= n; i++){    //
            if(dis[i] < min(dis[k], G[k][i])){   //假设k到i有路径,那么如果当前i点直接到1点的最大承载量,
                dis[i] = min(dis[k], G[k][i]);  //要比从1到k再从k到i这条路径的最大承载量小,那么我们就要更新,因为我们
            }                                  //求的是所有点到1点的最大承载量，当前1直接到i这条路径的最大承载量小了，我们就要更新

        }
    }
    return;
}

int main(){
    int t;
    int cas = 1;
    int n, m;  //n表示十字路口的个数,m表示街道的个数;
    int p1, p2, w;
    cin >> t;
    while(t--){
        printf("Scenario #%d:\n", cas++);
        scanf("%d%d", &n, &m);
        memset(G, 0, sizeof(G));  //初始化每条路的最大承载量都是0;
        for(int i = 0; i < m; i++){
            scanf("%d%d%d", &p1, &p2, &w);
            G[p1][p2] = w;
            G[p2][p1] = w;
        }
        dijkstra(n);
        printf("%d\n\n", dis[n]);
    }
    return 0;
}