题目1008：最短路径问题

题目描述：

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。

输入：

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t。n和m为0时输入结束。
(1<n<=1000, 0<m<100000, s != t)

输出：

输出 一行有两个数， 最短距离及其花费。

样例输入：

3 2
1 2 5 6
2 3 4 5
1 3
0 0

样例输出：

9 11

题目比较简单 用邻接矩阵+Dijkstra即可

#include <iostream>
#include <stdio.h>

using namespace std;
int data[1001][1001][2];
int dis[1001];
int fare[1001];
int points[1001];

bool isOver(int points[],int n){
    bool over = true;
    for(int i = 1; i <= n; i++){
        if(points[i] == 0){
            over = false;
            break;
        }
    }
    return over;
}

int main(){

    int n,m;

    scanf("%d %d",&n,&m);
    while(n != 0 && m != 0){
        for(int i = 0; i < 1001;i++){//初始化
            for(int j = 0; j < 1001;j++){
                data[i][j][0] = -1;
                data[i][j][1] = -1;
            }
            dis[i] = -1;
            fare[i] = -1;
            points[i] = 0;
        }
        for(int i = 0; i < m; i++){//读数据
            int point1,point2,distance,money;
            scanf("%d %d %d %d",&point1,&point2,&distance,&money);
            data[point1][point2][0] = distance;
            data[point1][point2][1] = money;
            data[point2][point1][0] = distance;
            data[point2][point1][1] = money;
        }

        int beginp,endp;
        scanf("%d %d",&beginp,&endp);

        dis[beginp] = 0;
        fare[beginp] = 0;
        points[beginp] = 1;
        for(int i = 1; i <= n; i++){
            if(data[beginp][i][0] != -1){
                dis[i] = data[beginp][i][0];
                fare[i] = data[beginp][i][1];
            }
        }

        while(!isOver(points,n)){
            int point = -1,anotherP= -1,minMoney,minDis = -1;
            for(int i = 1; i <= n;i++){
                if(points[i]==1){//已确定点
                    for(int j = 1; j <= n; j++){
                        if(points[j] == 0){//未确定点
                            if(data[i][j][0] ==-1)
                                continue;
                            if(minDis == -1 || minDis > data[i][j][0]){
                                minDis = data[i][j][0];
                                minMoney = data[i][j][1]+fare[i];
                                point = j;
                                anotherP = i;
                            }else if(minDis == data[i][j][0] && (fare[anotherP] + data[anotherP][point][1] > fare[i] + data[i][j][1])){//距离相同比较花费
                                minDis = data[i][j][0];
                                minMoney = data[i][j][1];
                                point = j;
                                anotherP = i;

                            }
                        }
                    }
                }
            }
            //
            points[point] = 1;
            dis[point] = dis[anotherP]+ data[anotherP][point][0];
            fare[point] = fare[anotherP] + data[anotherP][point][1];
            for(int i = 1; i <= n ;i++){
                if(points[i] == 0 &&data[point][i][0] != -1){
                    if(dis[i] == -1 || dis[i] > dis[point] +data[point][i][0]){
                        dis[i] = dis[point] + data[point][i][0];
                    }
                }
            }
        }

        printf("%d %d\n",dis[endp],fare[endp]);
        scanf("%d %d",&n,&m);
    }
    return 0;
}

题目1008：最短路径问题

原文：http://blog.csdn.net/mycomein/article/details/46712203