最短路径问题
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13771 Accepted Submission(s): 4234
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)
(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 2 1 2 5 6 2 3 4 5 1 3 0 0
Sample Output
9 11
Source
这道题我竟然在INFINITY这个点wa了许久,例子:#define INFINITY 0x7fffffff 那么INFINITY + any < 0 成为一个极小的数,那么结果自然就是错的
/****************************** * * acm: hdu-3790 * * title: 最短路径问题 * * time : 2014.8.29 * *******************************/ /* 思路: 1/2 例图 ① -> ② 2/2↘ ↙1/2 (距离/花费) ③ 可见,路径相等时,记录花费较少的即可,可直接用Dijkstra算法。 */ #include <stdio.h> #include <stdlib.h> #define MAXVEX 1001 #define MAXEDGE 100000 #define INFINITY 0x3fffffff //不能0x7fffffff => 加上1变成负数 成了一个很小的数 //构造边集 typedef struct Edge { int cost; //花费 int lenth; //长度 } Edge; Edge edges[MAXVEX][MAXVEX]; void CreateMGraph(int numVertexes, int numEdges) { int i, j; int a, b; int len, cost; for (i = 1; i <= numVertexes; i++) { for (j = 1; j <= numVertexes; j++) { edges[i][j].lenth = INFINITY; edges[i][j].cost = INFINITY; } } for (i = 0; i < numEdges; i++) { scanf("%d%d%d%d", &a, &b, &len, &cost); if (edges[a][b].lenth > len) { edges[a][b].lenth = len; edges[a][b].cost = cost; edges[b][a].lenth = len; edges[b][a].cost = cost; } else if (edges[a][b].lenth == len) { if (edges[a][b].cost > cost) { edges[a][b].cost = cost; edges[b][a].cost = cost; } } } /* 静态数据 edges[1][2].lenth = 5; edges[1][2].cost = 6; edges[2][1].lenth = 5; edges[2][1].cost = 6; edges[2][3].lenth = 4; edges[2][3].cost = 5; edges[3][2].lenth = 4; edges[3][2].cost = 5; */ } typedef int ShortPathTable[MAXVEX]; typedef int Cost[MAXVEX]; //Dijkstra算法 void ShortestPath_Dijkstra(int numVertexes, int numEdges, int v0, int vm, int *len, int *cost) { int v, w, k, min; int min_cost; ShortPathTable D; int final[MAXVEX]; //标记是否求得顶点v0-vm的路径 Cost C; *len = INFINITY; *cost = INFINITY; for (v = 1; v <= numVertexes; v++) { final[v] = 0; D[v] = edges[v0][v].lenth; C[v] = edges[v0][v].cost; } final[v0] = 1; //开始主循环,每次求得v0到某v顶点的最短路径 for (v = 2; v <= numVertexes; v++) { min = INFINITY; //当前所知离v0顶点的最近距离 min_cost = 0; for (w = 1; w <= numVertexes; w++) //寻找离v0最近的顶点 { if (!final[w] && D[w] < min) { k = w; min = D[w]; //w顶点离v0更近 min_cost = C[w]; } } final[k] = 1; if (k == vm) { *len = D[k]; *cost = C[k]; break; } for (w = 1; w <= numVertexes; w++) { if (!final[w] && (min+edges[k][w].lenth) < D[w]) //如果经过v顶点的路径比现在这条路径的长度短的话 { //说明找到了更短的路径,修改D[w],C[w] D[w] = min + edges[k][w].lenth; C[w] = min_cost + edges[k][w].cost; } else if (!final[w] && (min+edges[k][w].lenth) == D[w]) //如果经过v顶点的路径与现在这条路径相等 { if (min_cost + edges[k][w].cost < C[w]) //取花费较少的值赋值 { C[w] = min_cost + edges[k][w].cost; } } } } } int main() { int numVertexes = 3; int numEdges = 2; int start; int end; int len; int cost; while (scanf("%d%d", &numVertexes, &numEdges) && numVertexes != 0 && numVertexes != 0) { CreateMGraph(numVertexes, numEdges); scanf("%d%d", &start, &end); ShortestPath_Dijkstra(numVertexes, numEdges, start, end, &len, &cost); printf("%d %d\n", len, cost); } return 0; }