1030 Travel Plan （30 分)

2019-03-15

1030 Travel Plan （30 分)

A traveler’s map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (≤500) is the number of cities (and hence the cities are numbered from 0 to N−1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

1	City1 City2 Distance Cost

where the numbers are all integers no more than 500, and are separated by a space.

Output Specification:

For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

Sample Input:

Sample Output:

1	0 2 3 3 40

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

图的问题。城市之间有高速公路连接，给出高速的距离及费用，求指定两城市间最短路径，如有多条，则选择花费最小的那条。

分析

类似PAT.1003 Emergency，只不过那题选择征集到救援队最大的路径，这题选择花费最小的路径。从起点使用dijkstra算法即可。使用dis，cost数组记录到各城市的最短距离及最小花费。需要输出路径，使用pre[i]记录i的前驱结点。输出时利用栈即可。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include<sstream>

using namespace std;
int e[510][510];
int v[510][510];
int visit[510]={};
int dis[510];
int cost[510];
int pre[510];
int inf=999999999;
int main() {
    int n,m,s,d,c1,c2,co,di,minc=0,mind;
    cin>>n>>m>>s>>d;
    fill(e[0],e[0]+510*510,0);
    for(int i=0;i<m;i++){
        cin>>c1>>c2>>di>>co;
        e[c1][c2]=e[c2][c1]=di;
        v[c1][c2]=v[c2][c1]=co;
    }
    fill(visit,visit+510,0);
    fill(dis,dis+510,inf);
    dis[s]=0;
    cost[s]=0;
    pre[s]=s;
    for(int i=0;i<n;i++){
        int u=-1;
        mind=inf;
        for(int j=0;j<n;j++){
            if(!visit[j]&&dis[j]!=inf){
                if(dis[j]<mind){
                    u=j;
                    mind=dis[j];
                }
            }
        }
        if(u==-1)break;
        visit[u]=1;
        for(int j=0;j<n;j++){
            if(!visit[j]&&e[u][j]!=0){
                if(dis[j]>dis[u]+e[u][j]){
                    dis[j]=dis[u]+e[u][j];
                    pre[j]=u;
                    cost[j]=cost[u]+v[u][j];

                }else if(dis[j]==dis[u]+e[u][j]){
                    if(cost[j]>cost[u]+v[u][j]){
                        cost[j]=cost[u]+v[u][j];
                        pre[j]=u;
                    }
                }
            }
        }
    }
    vector<int> temp;
    int i=d;
    while(i!=s){
        temp.push_back(i);
        i=pre[i];
    }
    cout<<s;
    while(!temp.empty()){
        cout<<" "<<temp.back();
        temp.pop_back();
    }
    cout<<" "<<dis[d];
    cout<<" "<<cost[d];
    return 0;
}