pat 1003

来源：互联网发布：淘宝lol代练可靠吗编辑：程序博客网时间：2024/05/01 17:15

题意：

就是给你一张图，求一点到另外一点的花费最短时间的路径，如果存在多条，那么选取其中的价值最大的那一条

简单的搜索题，直接附代码(找自己的代码挺麻烦的，代码是自己的一个组员写的)

#include<stdio.h>#define MAXVER 501#define MAXEDG 12501typedef struct{  int vex[MAXVER];//teams in every city  int arc[MAXVER][MAXVER];//length of road  int N, M;}Mgraph;Mgraph g;int visited[MAXVER], c1, c2, count, max, min = 66235;void read(){  int i, j, r, c;  scanf("%d%d%d%d", &(g.N), &(g.M), &c1, &c2);  for(i = 0; i < g.N; i++)//the number of teams in i city    scanf("%d", &g.vex[i]);  for(i = 0; i < MAXVER; i++)//init, make every roads be 0    for(j = 0; j < MAXVER; j++){      g.arc[i][j] = 0;    }    for(i = 0; i < g.M; i++){//length of the road      scanf("%d%d%d", &r, &c, &j);      g.arc[r][c] = g.arc[c][r] = j;        }}void dfs(int now, int right, int team){  int i;  if(right > min)    return;  if(now == c2)  {    if(right == min){      count ++;      if(team >= max)        max = team;    }    if(right < min){      min = right;      count = 1;      max = team;    }    return ;  }  for(i = 0; i < g.N; i++){    if(!visited[i] && g.arc[now][i] > 0)  {      visited[i] = 1;      dfs(i, right + g.arc[now][i], team + g.vex[i]);      visited[i] = 0;    }  }}int main(){  //freopen("in.txt", "r", stdin);  read();//  if(c1 == c2)//    printf("1 g.vex[c1]");//  else{    visited[c1] = 1;    dfs(c1, 0, g.vex[c1]);    printf("%d %d", count, max); //  }  return 0;}

0 0