poj1797 Heavy Transportation dijkstra变形

Heavy Transportation

Time Limit: 3000MS
Memory Limit: 30000KTotal Submissions: 36131
Accepted: 9549

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

13 31 2 31 3 42 3 5

Sample Output

Scenario #1:4

题意：

火车从1点运到n点，每条路都有个限重，找一条路的所能承受的限重最大

相当于 ：给定一个无向图，求从1到n的路径上的最小值的最大值。

就是说，从1到n可能有多条路径，每条路径上都有一个权值最小的边，问这些边的最大值。

思路：dijkstra的变形  由原来的

if(!vis[j]&&dis[j]>dis[pos]+value[pos][j]){dis[j]=dis[pos]+value[pos][j];}

变为

if(!vis[j]&&dis[j]<min(dis[pos],value[pos][j])){dis[j]=min(dis[pos],value[pos][j]);}

同时不要忘记对输入的value数组的初始化赋值进行处理 ，是一个容易WA的点

#include<iostream>#include<cmath>#include<cstdio>#include<iomanip>#include<cstring>#include<stdlib.h>#include<string>#include<algorithm>#include<queue>using namespace std;#define MAXN 4500#define INF 0xFFFFFFFint value[MAXN][MAXN];/*保存的是边权值*/int dis[MAXN];/*保存源点到任意点之间的最短路*/int vis[MAXN];/*记录顶点是否没取过*/int n,m;void input(){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){value[i][j]=0;}}int u,v,s;for(int i=0;i<m;i++){scanf("%d%d%d",&u,&v,&s);if(s>value[u][v]){value[u][v]=s;value[v][u]=s;}}}void dijkstra(){memset(vis,false,sizeof vis);for(int i=1;i<=n;i++){dis[i]=0;}dis[1]=INF;for(int i=1;i<=n;i++){int pos=-1;for(int j=1;j<=n;j++){if(!vis[j]&&(pos==-1||dis[j]>dis[pos])){pos=j;}}vis[pos]=true;for(int j=1;j<=n;j++){if(!vis[j]&&dis[j]<min(dis[pos],value[pos][j])){dis[j]=min(dis[pos],value[pos][j]);}}}}int main(){int t;scanf("%d",&t);for(int k=1;k<=t;k++){scanf("%d%d",&n,&m);input();dijkstra();printf("Scenario #%d:\n",k);printf("%d\n\n",dis[n]);}return 0;}