Heavy Transportation

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 52728		Accepted: 13474

题目链接：http://poj.org/problem?id=1797

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

题意：

问从1号点到n号点所经路径的最小边权的最大值为多少。

 

题解：

其实本题也不是严格的dijkstra算法，只是利用了类似的贪心思想。

我们首先维护一个到当前点所有路径中的最小值，把它扔进优先队列里面，从优先队列里面每次取大值出来去更新与之相邻的点。

这里的正确性证明和dijkstra算法的证明类似，也就是说一个点去更新其它点后，不会被其他点又一次更新。

 

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1005;
int T;
int n,tot,m;
int head[N],vis[N],d[N];
struct Edge{
    int u,v,next,w;
}e[N*N<<1];
struct node{
    int d,u;
    bool operator < (const node &A)const{
        return d<A.d;
    }
};
void adde(int u,int v,int w){
    e[tot].v=v;e[tot].next=head[u];e[tot].w=w;head[u]=tot++;
}
void Dijkstra(int s){
    priority_queue <node> q;
    memset(d,0,sizeof(d));memset(vis,0,sizeof(vis));
    node now;d[s]=INF;
    now.d=INF;now.u=s;
    q.push(now);
    while(!q.empty()){
        node cur = q.top();q.pop();
        int u=cur.u;
        if(vis[u]) continue ;
        vis[u]=1;
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].v;
            if(d[v]<min(d[u],e[i].w)){
                d[v]=min(d[u],e[i].w);
                now.d=d[v];now.u=v;
                q.push(now);
            }
        }
    }
}
int main(){
    cin>>T;
    int cnt = 0,first=1;
    while(T--){
        cnt++;
        scanf("%d%d",&n,&m);
        memset(head,-1,sizeof(head));tot=0;
        for(int i=1;i<=m;i++){
            int u,v,c;
            scanf("%d%d%d",&u,&v,&c);
            adde(u,v,c);adde(v,u,c);
        }
        Dijkstra(1);
        printf("Scenario #%d:\n",cnt);
        cout<<d[n]<<endl;
        cout<<endl;
    }
    return 0;
}