1030. Travel Plan (30)

1. Travel Plan (30)
   时间限制
   400 ms
   内存限制
   65536 kB
   代码长度限制
   16000 B
   判题程序
   Standard
   作者
   CHEN, Yue

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:

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

4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20

Sample Output

0 2 3 3 40

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熟悉了此类题型的思路后，本题便十分简单了。

#include<iostream>#include<vector>#include<stack>using namespace std;vector<int> c[505];int path[505];int load[505][505];int cost[505][505];#define INF 100000000typedef struct node{    int dist,w;    bool visited;    node(){        visited=false;        dist=INF;        w=0;    }};node city[505];int n,m,c1,c2;int find(){    int find=-1,min=INF;    for(int i=0;i<n;i++){        if(city[i].dist<min&&city[i].visited==false)        {            min=city[i].dist;            find=i;        }    }    return find;}int main(){    freopen("in.txt","r",stdin);    while(scanf("%d%d%d%d",&n,&m,&c1,&c2)!=EOF)    {        for(int i=0;i<m;i++){            int t1,t2;            cin>>t1>>t2;            cin>>load[t2][t1]>>cost[t2][t1];            cost[t1][t2]=cost[t2][t1];            load[t1][t2]=load[t2][t1];            c[t1].push_back(t2);             c[t2].push_back(t1);        }        city[c1].dist=0;        city[c1].w=0;        int v=find();        while(v!=-1){            city[v].visited=true;            for(int i=0;i<c[v].size();i++)            {                int t=c[v][i];                if(city[t].visited==false){                    if(city[t].dist>city[v].dist+load[v][t])                    {                        city[t].dist=city[v].dist+load[v][t];                        city[t].w=cost[v][t]+city[v].w;                        path[t]=v;                    }                    else if(city[t].dist==city[v].dist+load[v][t])                    {                        if(city[t].w>cost[v][t]+city[v].w)                        {                            city[t].w=cost[v][t]+city[v].w;                            path[t]=v;                        }                    }                }            }            v=find();        }        int t=c2;        stack<int> pout;        pout.push(t);        while(t!=c1)        {            t=path[t];            pout.push(t);        }        while(!pout.empty())        {            printf("%d ",pout.top());            pout.pop();        }        printf("%d %d",city[c2].dist,city[c2].w);    }    return 0;}