POJ1797 Heavy Transpotation

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

刚开始的思路是把每条边的权值处理一下 用1000005-w作为权值，然后求最短路 再求路径上的最小的那个权值

但是实际上每一次都要尽量找最大的那个权值 而不是让和最大

所以正确的做法是改变一下松弛的条件【最短路题目的核心】

然而还是不太清楚要怎么改 参考了一下题解

dijkstra 和 sfpa都写了下

还有就是 最短路的题目要注意初始化

这道题用cin会T

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<cstring>
#include<queue>
#include<stack>
#define inf 0x3f3f3f3f

using namespace std;

const int maxn = 1005;
int t, n, m;
bool vis[maxn];
int p[maxn][maxn], d[maxn];

/*void dijkstra(int sec)
{
    int mmax, max_num;
    for(int i = 1; i <= n; i++ ){
        vis[i] = false;
        d[i] = p[sec][i];
    }
    vis[sec] = true;
    d[sec] = 0;
    for(int i = 1; i < n; i++){
        mmax = -inf;
        for(int j = 1; j <= n; j++){
            if(!vis[j] && d[j] > mmax){
                mmax = d[j];
                max_num = j;
            }
        }
        vis[max_num] = 1;
        for(int j = 1; j <= n; j++){
            if(!vis[j] && d[j] < min(p[max_num][j], d[max_num])){
                d[j] = min(p[max_num][j], d[max_num]);
            }
        }
    }
}*/

void spfa(int sec)
{
    queue <int> q;
    for(int i = 1; i <= n; i++){
        d[i] = -1;
        vis[i] = false;
    }

    d[sec] = inf;
    vis[sec] = true;
    q.push(sec);
    while(!q.empty()){
        int v = q.front();q.pop();
        vis[v] = false;
        for(int i = 1; i <= n; i++){
            int t = min(d[v], p[v][i]);
            if(d[i] < t){
                d[i] = t;
                if(!vis[i]){
                    vis[i] = true;
                    q.push(i);
                }
            }
        }
    }
}


int main()
{
    cin>>t;
    for(int cas = 1; cas <= t; cas++){
        memset(p, 0, sizeof(p));
        scanf("%d%d",&n,&m);
        for(int i = 0; i < m; i++){
            int a, b, c;
            scanf("%d%d%d",&a,&b,&c);
            p[a][b] = c;
            p[b][a] = c;
        }
        spfa(1);

        cout<<"Scenario #"<<cas<<":
";
        cout<<d[n]<<endl<<endl;
    }

    return 0;
}

dijkstra的思路：

做n-1次遍历 每次都找还没访问的节点中d[]最大的那个节点j【起点到这个节点的路径中 最小权值的边 比起点到其他节点的路径中最小权值的边的权值要大】

遍历这个结点的邻接点，做松弛操作 

如果这个邻接点 i 没有被访问过 如果他此时的d比   j 的 d 和 j 到 i 的边的权值的最小值要小   那么就要更新 i 的d【让起点到 i 的路径经过 j】

spfa的思路：

设置一个队列 将起点加入队列 每次从队列中取出队头    更新剩余结点

松弛条件和dijkstra类似

给边权值初始化为0， 这样他的权值比所有的d都要小， 也就不会赋值给任何的d了