poj 1797 Heavy Transportation

Heavy Transportation

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 14380		Accepted: 3785

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

#include<iostream>
using namespace std;
const int N = 1010;
int map[N][N];//用邻接矩阵表示的图
int d[N];//辅助数组，记录当前已经探测到的节点到各个节点的最小距离
int mark[N];//记录节点是否已经被探测到过

int p;//顶点数 
const int INF = INT_MAX;//设置一个极大值

void dijkstra(int v)
{
    int i, j;
    int mini;
    int pos = v;
    memset(mark, 0, sizeof(mark));
    for(i=1; i<=p; i++)//初始化都为极大值
    {
        d[i] = 0;
    }
    for(i=1; i<=p; i++)
    {
        d[i] = map[pos][i];
    }
    mark[v] = 1;
    d[v] = 0;
    for(i=1; i<p; i++)
    {
        mini = 0;
        for(j=1; j<=p; j++)
        {
            if(mark[j]!=1 && d[j]>mini)
            {
                mini = d[j];
                pos = j;
            }
        }
        mark[pos] = 1;
        
        for(j=1; j<=p; j++)//对于新加入的节点，重新扫描跟这个点相关的每一条边，进行松弛操作 
        {
            int temp = min(d[pos],map[pos][j]);
            if(temp>d[j] && !mark[j] && map[pos][j])
                d[j] = temp;
        }
    }
}

int main()
{
    int sene;
    int m;
    int x, y, weight;
    freopen("e:\\data.txt", "r", stdin);   
    freopen("e:\\out.txt", "w", stdout);
    cin>>sene;
    for(int s=1; s<=sene; s++)
    {
        cin>>p>>m;
        memset(map, 0, sizeof(map));
        memset(d, 0, sizeof(d));
        while(m--)
        {
            cin>>x>>y>>weight;
            map[x][y] = weight;
            map[y][x] = weight;
        }
        dijkstra(1);
        cout<<"Scenario #"<<s<<":"<<endl;
        cout<<d[p]<<endl<<endl;
    }
}

本来想用floyd算法尝试一下了，超时了

用d算法，稍微修改了一下就可以了