ZOJ2526 FatMouse and JavaBean II，Dijkstra算法+DFS

这题挺不错的，有一定难度。求到终点的最短路径我用Dijkstra，其路径更新条件可以看我的程序，如果到某个点有多个路径长度一样的最短路径，则选择豆子总数最多的。而求有多少条路径的长度是最短的，我没有想到什么好的方法，只能用DFS了。

/******************************************************************************* # Author : Neo Fung # Email : [e-mail] # Last modified: 2011-08-20 14:45 # Filename: ZOJ2526 FatMouse and JavaBean II.cpp # Description :  ******************************************************************************/// ZOJ2526 FatMouse and JavaBean II.cpp : Defines the entry point for the console application.//// #include "stdafx.h"#include <fstream>#include <stdio.h>#include <iostream>#include <memory.h>#include <stack>using namespace std;#define MAX_NUM 502#define INF 1<<30int mat[MAX_NUM][MAX_NUM];int bean[MAX_NUM];int dij[MAX_NUM];int visit[MAX_NUM];int last[MAX_NUM];int beansum[MAX_NUM];int DFS(int start,int end,int len,int n);int main(void){// ifstream cin("data.txt");int N,M,sou,des,a,b,len;while(cin>>N>>M>>sou>>des){for(int i=0;i<N;++i){for(int j=0;j<N;++j){mat[i][j]=INF;}mat[i][i]=0;last[i]=-1;dij[i]=INF;}memset(visit,0,sizeof(visit));memset(beansum,0,sizeof(beansum));for(int i=0;i<N;++i)cin>>bean[i];for(int i=0;i<M;++i){cin>>a>>b>>len;mat[a][b]=mat[b][a]=len;}int now=sou;dij[now]=0;visit[now]=1;beansum[now]=bean[now];for(int i=0;i<N;++i){int min(INF);for(int j=0;j<N;++j)if(!visit[j] && mat[now][j]<INF && dij[j]>=dij[now]+mat[now][j]){if(dij[j]>dij[now]+mat[now][j]){dij[j]=dij[now]+mat[now][j];beansum[j]=beansum[now]+bean[j];last[j]=now;}else if(beansum[j]<beansum[now]+bean[j]){beansum[j]=beansum[now]+bean[j];last[j]=now;}}for(int j=0;j<N;++j)if(!visit[j] && min>dij[j])min=dij[now=j];visit[now]=1;}stack<int> mystack;for (int i=des;i!=sou;i=last[i])mystack.push(i);memset(visit,0,sizeof(visit));int pathcount(DFS(sou,des,dij[des],N));// int pathcount(0);printf("%d %d\n",pathcount,beansum[des]);printf("%d",sou);while(!mystack.empty()){printf(" %d",mystack.top());mystack.pop();}printf("\n");}return 0;}int DFS(int start,int end,int len,int n){visit[start]=1;int result(0);if(start==end&&!len)result=1;else if(len<0)result=0;else{for(int i=0;i<n;++i){if(mat[start][i]<INF && !visit[i])result+=DFS(i,end,len-mat[start][i],n);}}visit[start]=0;return result;}