A Walk Through the Forest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7995 Accepted Submission(s): 2943
Problem 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.
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.
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
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
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
4
Source
题意：
从起点1到终点2，求这样的路的条数：如果从a点到2的距离大于从b点到2的距离，并且a能到b那么就从a走到b。问这样的路有几条。
代码：
//并不是求最短路的条数。先dijk出2到每个点的最短路，在记忆化搜索出路径数
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int MAX=;
int mp[][],dis[],vis[],path[];
int n,m;
void dijk()
{
for(int i=;i<=n;i++)
{
dis[i]=mp[][i];
vis[i]=;
}
vis[]=;
for(int i=;i<=n;i++)
{
int Min=MAX,sta=; //sta初始化一下，防止下面找不到j，而出现vis[sta]数组越界。
for(int j=;j<=n;j++)
{
if(!vis[j]&&dis[j]<Min)
{
Min=dis[j];
sta=j;
}
}
vis[sta]=;
for(int j=;j<=n;j++)
{
if(!vis[j]&&mp[sta][j]!=MAX&&dis[j]>dis[sta]+mp[sta][j])
dis[j]=dis[sta]+mp[sta][j];
}
}
}
int dfs(int x)
{
if(path[x]!=-) //记忆化
return path[x];
if(x==)
return ;
path[x]=;
for(int i=;i<=n;i++)
{
if(mp[x][i]!=MAX&&dis[x]>dis[i])
path[x]+=dfs(i);
}
return path[x];
}
int main()
{
int a,b,c;
while(scanf("%d",&n)&&n)
{
scanf("%d",&m);
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
mp[i][j]=i==j?:MAX;
for(int i=;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
mp[a][b]=mp[b][a]=min(mp[a][b],c);
}
dijk();
memset(path,-,sizeof(path));
int ans=dfs();
printf("%d\n",ans);
}
return ;
}
代码：