HDU3790 最短路径问题 【Dijkstra】

最短路径问题

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 13336    Accepted Submission(s): 4072

Problem Description

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

路径相同时应更新最小花费。

#include <stdio.h>#include <string.h>#define maxn 1002#define maxm 100002#define inf 0x7fffffffint head[maxn], id;struct Node{int to, cost, dist, next;} E[maxm << 1];bool vis[maxn];int dist[maxn], cost[maxn];void addEdge(int u, int v, int d, int c){E[id].to = v; E[id].cost = c; E[id].dist = d;E[id].next = head[u]; head[u] = id++;E[id].to = u; E[id].cost = c; E[id].dist = d;E[id].next = head[v]; head[v] = id++;}int getNext(int n){int i, tmp = inf, u = -1;for(i = 1; i <= n; ++i)if(!vis[i] && dist[i] < tmp){tmp = dist[i]; u = i;}return u;}void Dijkstra(int n, int s, int t){int u, v, i, count, tmpd, tmpc;memset(vis, 0, sizeof(vis));memset(dist, 0x7f, sizeof(dist));memset(cost, 0x7f, sizeof(cost));dist[s] = 0; u = s; cost[s] = 0;while(u != -1){for(i = head[u]; i != -1; i = E[i].next){tmpd = dist[u] + E[i].dist;tmpc = cost[u] + E[i].cost;v = E[i].to;if(tmpd < dist[v]){dist[v] = tmpd; cost[v] = tmpc;}else if(tmpd == dist[v] && tmpc < cost[v]){dist[v] = tmpd; cost[v] = tmpc;}}vis[u] = true;if(vis[t]) break;u = getNext(n);}}int main(){int n, m, i, u, v, c, d, s, t;while(scanf("%d%d", &n, &m), n || m){memset(head, -1, sizeof(head));for(i = id = 0; i < m; ++i){scanf("%d%d%d%d", &u, &v, &d, &c);addEdge(u, v, d, c);}scanf("%d%d", &s, &t);Dijkstra(n, s, t);printf("%d %d\n", dist[t], cost[t]);}return 0;}

2015.1.27更新

#include <stdio.h>#include <string.h>#define maxn 1002#define inf 0x3f3f3f3fint G[maxn][maxn], N;int Gp[maxn][maxn];struct Node {int d, p;} D[maxn];bool vis[maxn];int getNext() {int u = -1, d = inf, p = inf, i;for (i = 1; i <= N; ++i)if (!vis[i] && D[i].d < d) {u = i; d = D[i].d; p = D[i].p;} else if (!vis[i] && D[i].d == d && D[i].p > p) {u = i; p = D[i].p;}return u;}void Dijkstra(int s, int t) {int u = s, i;for (i = 1; i <= N; ++i) {vis[i] = 0; D[i].d = D[i].p = inf;}D[u].d = D[u].p = 0;while (u != t) {vis[u] = 1;for (i = 1; i <= N; ++i) {if (D[u].d + G[u][i] < D[i].d) {D[i].d = D[u].d + G[u][i];D[i].p = D[u].p + Gp[u][i];} else if (D[i].d == D[u].d + G[u][i] && D[i].p > D[u].p + Gp[u][i])D[i].p = D[u].p + Gp[u][i];}u = getNext();}printf("%d %d\n", D[t].d, D[t].p);}int main() {freopen("stdin.txt", "r", stdin);int M, u, v, d, p, s, t;int i, j;while (scanf("%d%d", &N, &M), N | M) {for (i = 1; i <= N; ++i)for (j = 1; j <= N; ++j)G[i][j] = Gp[i][j] = inf;while (M--) {scanf("%d%d%d%d", &u, &v, &d, &p);if (G[u][v] > d) {G[u][v] = G[v][u] = d; Gp[u][v] = Gp[v][u] = p;} else if (G[u][v] == d && Gp[u][v] > p)Gp[u][v] = Gp[v][u] = p;}scanf("%d%d", &s, &t);Dijkstra(s, t);}return 0;}