UVa 10917 A Walk Through the Forest
2015-10-27 11:11
309 查看
A Walk Through the Forest
Time Limit:1000MS Memory Limit:65536K
Total Submit:48 Accepted:15
Description
Jimmy experiences a lot of stress at work these days, especially since his
accident made working difficult. To relax after a hard day, he likes to walk
home. To make things even nicer, his office is on one side of a forest, and his
house is on the other. A nice walk through the forest, seeing the birds and
chipmunks is quite enjoyable.
The forest is beautiful, and Jimmy wants to take a different route everyday. He
also wants to get home before dark, so he always takes a path to make progress
towards his house. He considers taking a path from A to B to be progress if
there exists a route from B to his home that is shorter than any possible route
from A. Calculate how many different routes through the forest Jimmy might
take.
Input
Input 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.
Output
For 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
【思路】
最短路+记忆化搜索。
SPFA预处理出每个点到达home的最短距离d,然后沿着d变小的边记忆化搜索路径条数。
【代码】
Time Limit:1000MS Memory Limit:65536K
Total Submit:48 Accepted:15
Description
Jimmy experiences a lot of stress at work these days, especially since his
accident made working difficult. To relax after a hard day, he likes to walk
home. To make things even nicer, his office is on one side of a forest, and his
house is on the other. A nice walk through the forest, seeing the birds and
chipmunks is quite enjoyable.
The forest is beautiful, and Jimmy wants to take a different route everyday. He
also wants to get home before dark, so he always takes a path to make progress
towards his house. He considers taking a path from A to B to be progress if
there exists a route from B to his home that is shorter than any possible route
from A. Calculate how many different routes through the forest Jimmy might
take.
Input
Input 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.
Output
For 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
【思路】
最短路+记忆化搜索。
SPFA预处理出每个点到达home的最短距离d,然后沿着d变小的边记忆化搜索路径条数。
【代码】
#include<cstdio> #include<queue> #include<vector> #include<cstring> using namespace std; const int maxn = 1000+10; const int INF=1<<30; struct Edge{ int u,v,w,next; }e[2*maxn*maxn]; int en,front[maxn]; int n,m; inline void AddEdge(int u,int v,int w) { en++; e[en].v=v; e[en].w=w; e[en].next=front[u]; front[u]=en; } int d[maxn]; void SPFA(int s) { int inq[maxn]; queue<int> q; memset(inq,0,sizeof(inq)); for(int i=1;i<=n;i++) d[i]=INF; d[s]=0; inq[s]=1; q.push(s); while(!q.empty()) { int u=q.front(); q.pop(); inq[u]=0; for(int i=front[u];i>=0;i=e[i].next) { int v=e[i].v,w=e[i].w; if(d[v]>d[u]+w) { d[v]=d[u]+w; if(!inq[v]) { inq[v]=1; q.push(v); } } } } } int f[maxn]; int dp(int u) { int& ans=f[u]; if(ans>=0) return ans; if(u==2) return ans=1; ans=0; for(int i=front[u];i>=0;i=e[i].next) { int v=e[i].v; if(d[v]<d[u]) ans += dp(v); //只沿着d更小的走 } return ans; } int main() { while(scanf("%d%d",&n,&m)==2) { en=-1; memset(front,-1,sizeof(front)); int u,v,w; for(int i=0;i<m;i++) { scanf("%d%d%d",&u,&v,&w); AddEdge(u,v,w); AddEdge(v,u,w); } SPFA(2); memset(f,-1,sizeof(f)); printf("%d\n",dp(1)); } return 0; }
相关文章推荐
- 2. Python中的序列——列表和元组
- centos6.5 安装svn可视化管理工具 if.svnadmin
- FIR IIR 数字滤波器 C++实现
- jsp中写java代码的方式
- POJ 2676 锻炼码力:数独,精确覆盖的DLX
- Android Dialog设置透明背景以及位置
- windows7批量设置文件权限命令参数详解
- Mongodb获取附近的人
- js用fuction定义构造函数
- R in Action 学习笔记-边学边查
- OC --- block
- iOS 第三方开源库的吐槽和备忘
- 热电制冷器的优缺点
- tomcat监控脚本
- JS基于Ajax实现的网页Loading效果代码
- 10-5 脚本编程之八 脚本完成磁盘分区格式化
- 几种经典常用加密算法
- 【bzoj3943】 [Usaco2015 Feb]SuperBull 最大生成树
- Javascript中的window.event.keyCode使用介绍
- spring email.properties