标签：direction clu quit beautiful cti following ever fine dfs
InputInput contains several test cases followed by a line containing 0. Jimmy has numbered each intersection or joining of paths starting with 1. His office is numbered 1, and his house is numbered 2. The first line of each test case gives the number of intersections N, 1 < N ≤ 1000, and the number of paths M. The following M lines each contain a pair of intersections a b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between intersection a and a different intersection b. Jimmy may walk a path any direction he chooses. There is at most one path between any pair of intersections.
OutputFor each test case, output a single integer indicating the number of different routes through the forest. You may assume that this number does not exceed 2147483647
Sample Input
5 6 1 3 2 1 4 2 3 4 3 1 5 12 4 2 34 5 2 24 7 8 1 3 1 1 4 1 3 7 1 7 4 1 7 5 1 6 7 1 5 2 1 6 2 1 0
Sample Output
2 4
题目大意：我们要从1到达2 问我们有多少种走法。但是我们不能越走越远，就是我们选择走这条路，只能距离2这个点越来越近。
#include<iostream> #include<cstdio> #include<cstring> #include<ctime> #include<cstdlib> #include<algorithm> #include<cmath> #include<string> #include<queue> #include<vector> #include<stack> #include<list> #include<set> #include<map> using namespace std; #define P pair<int ,int > const int maxn=1000+10; const int INF = 0x3f3f3f3f; int Lext[maxn],Next[maxn*200],To[maxn*200],Len[maxn*200],dis[maxn],cost[maxn]; int cnt; void add(int u,int v,int w) { Next[++cnt]=Lext[u]; Lext[u]=cnt; To[cnt]=v; Len[cnt]=w; } void init() { cnt=0; memset(Lext,-1,sizeof(Lext)); memset(dis,INF,sizeof(dis)); memset(cost,0,sizeof(cost)); } void dij(int st) { dis[st]=0; priority_queue<P,vector<P >, greater<P > >q; q.push(P(0,st)); while(!q.empty()) { P temp=q.top(); q.pop(); int x=temp.second; for(int i=Lext[x]; i!=-1; i=Next[i]) { int y=To[i]; int d=Len[i]; if(dis[y]>dis[x]+d) { dis[y]=dis[x]+d; q.push(P(dis[y],y)); } } } } int dfs(int st) { //cost数组用来存一共有多少走法 if(st==2) return 1; if(cost[st]) return cost[st]; for(int i=Lext[st]; i!=-1; i=Next[i]) { int y=To[i]; if(dis[y]<dis[st]) cost[st]+=dfs(y); } return cost[st]; } int main() { int n,m,u,v,w; while(scanf("%d",&n),n) { init(); scanf("%d",&m); for(int i=0; i<m; i++) { scanf("%d %d %d",&u,&v,&w); add(u,v,w); add(v,u,w); } dij(2); cout<<dfs(1)<<endl; } }
HDU - 1142 A Walk Through the Forest (最短路)
标签：direction clu quit beautiful cti following ever fine dfs
原文地址：https://www.cnblogs.com/MengX/p/9062847.html
