HDU 3790

最短路径问题

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14009    Accepted Submission(s): 4304

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

 

加了输入挂，果然快了一倍。然后个人还是不够严谨

 

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 1010
int d[N][N],cost[N][N],vis[N],dist[N],cst[N];
int n,m,a,b,dd,p,s,t;
const int inf = 0X700000;
int in() //输入挂
{
    int r = 0;
    char ch;
    ch = getchar();
    while(ch < '0' || ch > '9') ch = getchar();
    while(ch >= '0' && ch <= '9')
    {
        r = r*10 + ch-'0';
        ch = getchar();
    }
    return r;
}
void init()
{
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++)
        {
            d[i][j] = cost[i][j] = inf;
            d[j][i] = cost[j][i] = inf;
        }
        vis[i] = 0;
        dist[i] = cst[i] = inf;
    }
}
void input()
{
    while(m--)
    {
        //scanf("%d %d %d %d",&a,&b,&dd, &p);
        a = in();
        b = in();
        dd = in();
        p = in();
        if(dd < d[a][b]){
            d[a][b]= d[b][a] = dd;
            cost[a][b] = cost[b][a] = p;
        }else if(dd == d[a][b] && p < cost[a][b])
        {
            cost[a][b] = cost[b][a] = p;
        }
        //d[a][b] = d[b][a] = min(d[a][b],dd);  这行错误 = =！！果然还是不够严谨
        //cost[a][b] = cost[b][a] = min(cost[a][b], p);
    }
    s = in(); t = in();
}

void dij()
{
    for(int i = 1; i <= n; i++){
        dist[i] = d[s][i];
        cst[i] = cost[s][i];
    }
    vis[s] = 1;
    dist[s] = 0;
    cst[s] = 0;
    for(int i = 2; i <= n; i++)
    {
        int mn = inf;
        int pos = 1;
        for(int j = 1; j <= n; j++)
            if(vis[j] == 0 && dist[j] < mn) {pos = j; mn = dist[j];}
        vis[pos] = 1;
        for(int j = 1; j <= n; j++)
        {
            if(vis[j] == 0 && d[pos][j] < inf)
            {
                int newd = dist[pos] + d[pos][j];
                int newc = cst[pos] + cost[pos][j];
                if(newd < dist[j]) {
                        dist[j] = newd;
                        cst[j] = newc;
                }else if(newd == dist[j] && newc < cst[j])
                    cst[j] = newc;
            }
        }
    }

}
int main()
{
    while(scanf("%d %d", &n, &m) && (m||n))
    {
        init();
        input();
        dij();
        printf("%d %d\n", dist[t], cst[t]);
    }
    return 0;
}