体会：这个题做了好久，还是参考了别人的代码。不过体会到不能随便用dfs了。
关键：用DIJKTRA,不过不知道具体是怎么样，，不过感觉这个很靠谱
#include<iostream>
#include<stdlib.h>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<stack>
#include<math.h>
#include<stdlib.h>
#include<list>
#include<vector>
using namespace std;
#define N 1001
int map[N][N];
int dis[N];
bool vist[N];
int n;
int m;
int ans;
int g[N];
void dfs(int k)
{
cout<<k<<endl;
int i,j;
vist[k]=true;
for (i=1;i<=n;i++)
{
if (!vist[i]&&map[i][k])
dfs(i);
if (map[i][k]&&dis[k]>dis[i]+map[i][k])
dis[k]=dis[i]+map[i][k];
}
}
void put()
{
int i,j,k;
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
cout<<map[i][j]<<' ';
cout<<endl;
}
for (i=1;i<=n;i++)
cout<<dis[i]<<' ';
cout<<endl;
}
void make(int k)
{
int i,j;
if (k==2)
{
ans++;
return;
}
for (i=1;i<=n;i++)
{
if (k!=i&&dis[i]<dis[k])
make(i);
}
}
void dij()
{
int i,j,k;
for (k=1;k<=n;k++)
{
int big=100000000;
for (i=1;i<=n;i++)
{
if (!vist[i]&&dis[i]<big)
big=dis[i],j=i;
}
//cout<<j<<endl;
vist[j]=true;
for (i=1;i<=n;i++)
{
if (!vist[i]&&map[i][j]&&dis[i]>map[i][j]+dis[j])
dis[i]=map[i][j]+dis[j];
}
}
}
int get(int k) //不能随便用dfs
{
int i,j;
if (g[k])
return g[k];
if (k==2)
return 1;
for (i=1;i<=n;i++)
{
if (map[k][i]&&dis[k]>dis[i])
g[k]+=get(i);
}
return g[k];
}
int main()
{
int i,j,k;
while (cin>>n)
{
if (n==0)
break;
cin>>m;
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
map[i][j]=0;
while (m--)
{
cin>>i>>j>>k;
map[i][j]=map[j][i]=k;
}
for (i=1;i<=n;i++)
dis[i]=100000000;
dis[2]=0;
memset(vist,false,sizeof(vist));
//dfs(1); //介个不行啊，，，这只能找到一点到另一点之间的最短距离，，不好用。
dij();
ans=0;
//make(1);
//put();
memset(g,0,sizeof(g));
ans=get(1);
cout<<ans<<endl;
}
}
A Walk Through the Forest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2856 Accepted Submission(s): 1034
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.
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
Source
University of Waterloo Local Contest 2005.09.24
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