pat 1003

          此题为图论算法的Dijskal算法，题目大意是给定一个图和各个节点的值和各个边长，要求求出起点到终点最短距离的条数，经过各个最短距离中各个点的最大顶点和，怎么想？怎么做？感觉对Dijiskal算法还不是真真的了解，首先，来说说最短路径的条数，如何来求出最短路径的条数？是先求出最短的一点，然后再松弛最短路径的条数，这样最后就可以求出最短路径的条数，那么如何来求经过各个最短距离中各个点的最大顶点和，这个也用松弛的办法，这题和pat1030比较相像，基本思路是一样的，先附代码如下：

#include<iostream>#include<cstdio>using namespace std;#define MAX 0x3fffffffint N,M,C1,C2;int teams[510];int map[510][510];int path[510];int path_teams[510];int dis[510];bool flag[510];void Dijskal(){flag[C1] = true;int i,j;for(j = 1;j < N;j++){int _min = MAX;int index;for(i = 0;i < N;i++){if(dis[i] < _min && flag[i] == false){_min = dis[i];index = i;}}flag[index] = true;for(i = 0;i < N;i++){if(flag[i] == false){if(dis[i] > dis[index] + map[i][index]){dis[i] = dis[index] + map[i][index];path[i] = path[index];path_teams[i] = path_teams[index] + teams[i];}else if(dis[i] == dis[index] +map[i][index]){path[i] = path[i] + path[index];if(path_teams[i] < path_teams[index] + teams[i])path_teams[i] = path_teams[index] + teams[i];}}}}}int main(){scanf("%d %d",&N,&M);scanf("%d %d",&C1,&C2);int i,j;for(i = 0;i < N;i++)scanf("%d",teams + i);for(i = 0;i < N;i++){dis[i] = MAX;flag[i] = false;path[i] = 0;path_teams[i] = 0;for(j = 0;j < N;j++){map[i][j] = MAX;}}int is,it,idis;for(i = 0;i < M;i++){scanf("%d %d %d",&is,&it,&idis);map[is][it] = idis;map[it][is] = idis;}for(i = 0;i < N;i++){dis[i] = map[i][C1];}for(i = 0;i < N;i++){if(dis[i] != MAX){path[i] = 1;path_teams[i] = teams[C1] + teams[i];}}if(C1 != C2){Dijskal();cout<<path[C2]<<" "<<path_teams[C2]<<endl;}else{cout<<"1 "<<teams[C1]<<endl;}return 0;}