Heavy Transportation

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

解题思路：最短路变形题，求两点之间经过的路的最大值。普通的最短路是求最短的路径的和，因此只要把求和改成求最大即可。详细见代码！

Dijkstra邻接矩阵法

#include<iostream>
#include<deque>
#include<memory.h>
#include<stdio.h>
#include<string>
#include<algorithm>
#include<vector>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#define inf 1<<29
#define MAXV 1005
using namespace std;


#define MAX 0x3f3f3f3f


int rect[MAXV][MAXV];
int n,m;
int tt;


void dijkstra(int s,int e){
    int i,j,f,v;

    //之前保存的是最短路径，现在保存的是最窄的路的宽度
    int d[MAXV];

    bool vis[MAXV];

    for(i=1;i<=n;i++){
        vis[i]=0;
        d[i]=rect[s][i];
    }

    for(i=1;i<=n;i++){
        //之前是取出路径最小的点来更新，现在是取出当前最宽的点，这里要稍作思考。
        f=-1;
        for(j=1;j<=n;j++)
            if(!vis[j]&&d[j]>f){
                v=j;
                f=d[j];
            }
        vis[v]=1;

        for(j=1;j<=n;j++)
            //之前是更新为更短的路径，现在是更新为更窄的路，就看看当前的窄，还是新加入的窄。
            if(!vis[j]&&d[j]<min(d[v],rect[v][j]))
                d[j]=min(d[v],rect[v][j]);
    }

    printf("Scenario #%d:\n",tt++);
    printf("%d\n\n",d[e]);

}


int main()
{
    tt=1;
    int t;
    scanf("%d",&t);
    int a,b,c;
    while(t--){

        //之前是初始化为最大，这里因为要求最宽的路，因此初始化为0
        memset(rect,0,sizeof(rect));
        
        scanf("%d%d",&n,&m);

        for(int i=0;i<m;i++){
            scanf("%d%d%d",&a,&b,&c);

            rect[a][b]=c;
            rect[b][a]=c;

        }


        dijkstra(1,n);
    }
    return 0;
}

SPFA邻接矩阵

#include<iostream>
#include<deque>
#include<memory.h>
#include<stdio.h>
#include<string>
#include<algorithm>
#include<vector>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#define inf 1<<29
#define MAXV 1005
using namespace std;


#define MAX 0x3f3f3f3f


int rect[MAXV][MAXV];
int n,m;
int tt;

int time[MAXV];
int d[MAXV];
int vis[MAXV];

bool spfa(int s,int e){
    
    queue<int> que;
    memset(time,0,sizeof(time));
    
    for(int i=1;i<=n;i++)d[i]=0;
    
    memset(vis,0,sizeof(vis));
    
    que.push(e);
    d[e]=MAX;//初始化为最宽
    time[e]++;
    vis[e]=1;
    
    while(!que.empty()){
        int tp=que.front();
        que.pop();
        
        vis[tp]=0;
        
        for(int i=1;i<=n;i++){
            //同上描述，看看有没有更窄的
            if(d[i]<min(d[tp],rect[tp][i])){
                d[i]=min(d[tp],rect[tp][i]);
                
                if(!vis[i]){
                    que.push(i);
                    vis[i]=1;
                    
                    if(++time[i]>=n)
                        return 1;
                }
                
            }
            
        }
        
    }
    
    printf("Scenario #%d:\n%d\n\n", tt++, d[s]);
    
    
    return 0;
    
}


int main()
{
    tt=1;
    int t;
    scanf("%d",&t);
    int a,b,c;
    while(t--){
        
        memset(rect,0,sizeof(rect));
        scanf("%d%d",&n,&m);
        
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&a,&b,&c);
            
            rect[a][b]=c;
            rect[b][a]=c;
            
        }
        
        
        spfa(1,n);
    }
    return 0;
}