hdu 1142 A Walk Through the Forest（spfa+深搜）

来源：互联网发布：兰州知豆招聘信息编辑：程序博客网时间：2024/06/10 09:48

http://blog.acmj1991.com/?p=785

题意：a->b的条件是：a到终点的最短距离比b到终点的最短距离要大，问起点到终点有多少条路能走？

思路：用spfa求出终点为源点到其他点的最短距离，然后依据题目条件进行深搜

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;#define maxN 1010#define inf 0x0fffffffint num,head[maxN],vist[maxN],dis[maxN],dp[maxN];struct node{int v,w,next;}po[maxN*maxN];void init(){num=0;memset(head,-1,sizeof(head));memset(vist,0,sizeof(vist));memset(dp,-1,sizeof(dp));}void add(int u,int v,int w){po[num].v=v,po[num].w=w;po[num].next=head[u],head[u]=num++;po[num].v=u,po[num].w=w;po[num].next=head[v],head[v]=num++;}void spfa(int n,int s){for(int i=0;i<=n;i++)dis[i]=inf;dis[s]=0,vist[s]=1;int stack[200*maxN],hea=0,tail=0;stack[tail++]=s;while(hea!=tail){int u=stack[hea++];vist[u]=0;for(int i=head[u];i!=-1;i=po[i].next){int v=po[i].v;if(dis[u]+po[i].w<dis[v]){dis[v]=dis[u]+po[i].w;if(!vist[v]){vist[v]=1;stack[tail++]=v;}}}}}int dfs(int u){if(u==2)return 1;if(dp[u]!=-1)return dp[u];else dp[u]=0;  for(int i=head[u];i!=-1;i=po[i].next){int v=po[i].v;if(dis[v]<dis[u]&&dis[v]<=dis[1]){dp[u]+=dfs(v);}}return dp[u];}int main(){int n,m;while(scanf("%d",&n)&&n){int u,v,w;init();scanf("%d",&m);while(m--){scanf("%d%d%d",&u,&v,&w);add(u,v,w);}spfa(n,2);dfs(1);printf("%d\n",dp[1]);}}