HDU 3790 简单最短路径问题（dijkstra+双重权值）

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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 21 2 5 62 3 4 51 30 0

Sample Output

9 11

思路：这也是这道题的重点，去掉重复的边，如果边相同，选择金钱比较少的边

上代码

#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#include<stack>#include<queue>#include<vector>#include<cstdio>#include<string>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define Max 1010int n, pp[Max][Max], M[Max][Max], d[Max], cost[Max];bool visit[Max];vector<pair<int, int> >G[Max];void init(int nn){for (int i = 1; i <= nn; i++){for (int j = 1; j <= nn; j++){M[i][j] = inf;pp[i][j] = inf;}d[i] = inf;cost[i] = inf;}memset(visit, false, sizeof(visit));}void dijkstra(int s, int t){int minv;d[s] = 0, cost[s] = 0;while (1){minv = inf;int u = -1;for (int i = 1; i <= n; i++){if (minv>d[i] && visit[i] == false){u = i;minv = d[i];}}if (u == -1)break;visit[u] = true;for (int i = 1; i <= n; i++){if (visit[i])continue;if (d[i] == d[u] + M[u][i] && M[u][i] != inf&&cost[i]>cost[u] + pp[u][i])cost[i] = cost[u] + pp[u][i];if (d[i]>d[u] + M[u][i]){d[i] = d[u] + M[u][i];cost[i] = cost[u] + pp[u][i];}}}cout << d[t] << " " << cost[t] << endl;}int main(){//freopen("Text1.txt","r",stdin);int m, u, v, d, p, s, t;while (scanf("%d%d", &n, &m) != EOF){if (n == 0 && m == 0)break;init(n);for (int i = 0; i<m; i++){scanf("%d%d%d%d", &u, &v, &d, &p);if (M[u][v]>d){M[u][v] = M[v][u] = d;pp[u][v] = pp[v][u] = p;}if (M[u][v] == d&&pp[u][v] > p)//边相同选择最小金钱的边，否则去掉pp[u][v] = pp[v][u] = p;}scanf("%d%d", &s, &t);dijkstra(s, t);}return 0;}

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