PAT A1030 Travel Plan (30分)

題目鏈接：https://pintia.cn/problem-sets/994805342720868352/problems/994805464397627392

題目描述
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.

輸入
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.

輸出
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.

樣例輸入
4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20

樣例輸出
0 2 3 3 40

代碼

#include <bits/stdc++.h>
using namespace std;

//n<=500可用鄰接矩陣表示
const int maxn = 510;
const int INF = 0x3f3f3f3f;
int n,m;
int G[maxn][maxn]; //存距離
int cost[maxn][maxn]; //存花費
int d[maxn], c[maxn]; //存最短距離和最短花費
int pre[maxn]; //pre[u]=v v是u在最短路徑中的前一個節點
bool vis[maxn];

//d[] s到每一點的最短距離
void Dijkstra(int s)
{
    vis[s] = true;
    fill(vis, vis + maxn, false);
    fill(d, d + maxn, INF);
    fill(c, c + maxn, INF);
    d[s] = 0;
    c[s] = 0;
    for(int i = 0; i < n; ++i) pre[i] = i;
    for(int i = 0; i < n; ++i){
        if(vis[i]) continue;
        //找未訪問節點中d[]最小的
        int minn = INF, u = -1;
        for(int j = 0; j < n; ++j){
            if(!vis[j] && d[j] < minn){
                minn = d[j];
                u = j;
            }
        }
        if(u == -1) return;
        vis[u] = true;
        for(int v = 0; v < n; ++v){
            if(!vis[v] && G[u][v] != INF){
                if(d[u] + G[u][v] < d[v]){
                    d[v] = d[u] + G[u][v];
                    c[v] = c[u] + cost[u][v];
                    pre[v] = u;
                }
                else if(d[u] + G[u][v] == d[v] && c[u] + cost[u][v] < c[v]){
                    c[v] = c[u] + cost[u][v];
                    pre[v] = u;
                }
            }
        }
    }
}
//遞歸求最短路徑
void DFS(int s, int v)
{
    if(v == s){
        printf("%d ",v);
        return;
    }
    DFS(s, pre[v]);
    printf("%d ",v);
}

int main()
{
    int s,e;
    fill(G[0], G[0] + maxn * maxn, INF);//首地址，尾地址，初始化的值
    //fill(cost[0], cost[0] + maxn * maxn, INF); //cost數組其實不必初始化
    scanf("%d%d%d%d",&n,&m,&s,&e);
    for(int i = 0; i < m; ++i){
        int a,b,dis,cos;
        scanf("%d%d%d%d",&a,&b,&dis,&cos);
        G[a][b] = G[b][a] = dis;
        cost[a][b] = cost[b][a] = cos;
    }
    Dijkstra(s);
    DFS(s, e);
    printf("%d %d",d[e], c[e]);
    return 0;
}