poj 1797 路径问题 dijkstra改写

题目：

Heavy Transportation

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 31194		Accepted: 8279

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

给n个点m条双向边，起点为1，终点为2，求从起点到终点所有路径中每条路径上各边最大权值的最小值。

分析：

可以用最短路做，需要对现有算法稍作改写。

既然是求最大值，松弛时就应该往大了改，可以先找出到这个点的最大负载量已经确定的点，松弛过程可以确保这个值不会继续变大，从这个点出发更新所有能更新的顶点。

注意输出格式，多一个换行。

听说也可以用最大增广路来做？学了之后来补

代码：

1.Floyd算法TLE

#include <stdio.h>
#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>
#include <math.h>
#include <limits.h>
using namespace std;
const int maxn=1009;
const int inf=100000000;

int n,m,t;
int floyd[maxn][maxn];

int main(){
	//求从1到n所有通路中最小跳的最大值 
	scanf("%d",&t);
	int a,b,c;
	for(int cas=1;cas<=t;++cas){
	    scanf("%d%d",&n,&m);
		for(int i=0;i<n;++i){
			for(int j=0;j<n;++j){
				floyd[i][j]=0;
			}
		}
	    for(int i=0;i<m;i++){
	    	scanf("%d%d%d",&a,&b,&c);
	    	if(c>floyd[a][b]) floyd[a][b]=c;
	    }
	    for(int k=1;k<=n;k++)
	        for(int i=1;i<=n;i++)
	            for(int j=1;j<=n;j++)
	                floyd[i][j]=max(min(floyd[i][k],floyd[k][j]),floyd[i][j]);
	    printf("Scenario #%d:\n%d\n",cas,floyd[1][n]);		
	}
    return 0;
}

2.dijkstra+松弛改写

#include <stdio.h>
#include <iostream>
#include <queue>
#include <vector>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <limits.h>
using namespace std;
const int maxn=1009;
const int inf=(INT_MAX)/3;

int n,m,t,d[maxn];//d数组存储从起点到i点的所有道路最小承重量的最大值
int g[maxn][maxn],used[maxn]; 

void dijkstra(int s){
	for(int i=1;i<=n;++i)
		d[i]=g[s][i];		//初值不可全置零 
	fill(used,used+n+1,0);
	d[s]=0;
	while(1){
		int v=-1;
		for(int i=1;i<=n;++i){
			if(!used[i]&&(v==-1||d[i]>d[v])) v=i;//找出最大负载量已经确定的点
		}
		if(v==-1) break;
		used[v]=1;
		for(int i=1;i<=n;++i){
			d[i]=max(d[i],min(d[v],g[v][i]));//和新发现通路最小承重量作比较，取较大值 
		}
	}
}

int main(){
	scanf("%d",&t);
	int a,b,c;
	for(int cas=1;cas<=t;++cas){
	    scanf("%d%d",&n,&m);
		memset(g,0,sizeof(g));
	    for(int i=0;i<m;i++){
	    	scanf("%d%d%d",&a,&b,&c);
	    	if(c>g[a][b]) g[a][b]=g[b][a]=c;
	    }
	    dijkstra(1);
	    printf("Scenario #%d:\n%d\n\n",cas,d[n]);
	}
    return 0;
}