HDU3790

最短路径问题

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23797    Accepted Submission(s): 7091

Problem Description

给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。

 

Input

输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点。n和m为0时输入结束。
(1<n<=1000, 0<m<100000, s != t)

 

Output

输出 一行有两个数， 最短距离及其花费。

 

Sample Input

3 2

1 2 5 6

2 3 4 5

1 3

0 0

 

Sample Output

9 11

 

//2016.4.21
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 1005;
const int inf = 9999999;
int Tu_dist[maxn][maxn], Tu_pay[maxn][maxn], dis[maxn], pay[maxn], book[maxn];
int n, m;
void dijkstra(int s);

int main()
{
    while(cin>>n>>m&&n&&m)
    {
        for(int i = 0; i <= n; i++)
            for(int j = 0; j <= n; j++)
            {
                Tu_dist[i][j] =inf;
                Tu_pay[i][j] = inf;
            }

        int a, b, d, p, s, t;
        while(m--)
        {
            scanf("%d%d%d%d", &a, &b, &d, &p);
            if(d<Tu_dist[a][b])
            {
                Tu_dist[a][b] = Tu_dist[b][a] = d;
                Tu_pay[a][b] = Tu_pay[b][a] = p;
            }
        }
        scanf("%d%d", &s, &t);

        dijkstra(s);
        printf("%d %d
", dis[t], pay[t]);
    }
    return 0;
}

void dijkstra(int s)
{
    int mindist, u, v;
    for(int i = 1; i <= n; i++)
    {
        dis[i] = Tu_dist[s][i];
        pay[i] = Tu_pay[s][i];
    }

    memset(book, 0, sizeof(book));
    book[s] = 1;
    dis[s] = 0;
    pay[s] = 0;

    for(int i = 1; i < n; i++)
    {
        mindist = inf;
        for(int j = 1; j <= n; j++)
        {
            if(book[j]==0&&dis[j]<mindist)
            {
                mindist = dis[j];
                u = j;
            }
        }
        book[u] = 1;
        for(int v = 1; v <= n; v++)
        {
            if(!book[v] && Tu_dist[u][v]<inf)
            {
                if(dis[v]>dis[u]+Tu_dist[u][v])
                {
                    dis[v] = dis[u]+Tu_dist[u][v];
                    pay[v] = pay[u]+Tu_pay[u][v];
                }
                else if(dis[v]==(dis[u]+Tu_dist[u][v]))
                    pay[v] = (pay[u]+Tu_pay[u][v])<pay[v]?pay[u]+Tu_pay[u][v]:pay[v];
            }
        }
    }
}