ACM: 图论题+记忆DP poj 2662 (边…

                             A Walk Through the Forest

Description

Jimmy experiences alot of stress at work these days, especially since his accidentmade working difficult. To relax after a hard day, he likes to walkhome. To make things even nicer, his office is on one side of aforest, and his house is on the other. A nice walk through theforest, seeing the birds and chipmunks is quite enjoyable.
The forest is beautiful, and Jimmy wants to take a different routeeveryday. He also wants to get home before dark, so he always takesa path to make progress towards his house. He considers taking apath from A to B to be progress if there exists a route from B tohis home that is shorter than any possible route from A. Calculatehow many different routes through the forest Jimmy mighttake.

Input

Input containsseveral test cases followed by a line containing 0. Jimmy hasnumbered each intersection or joining of paths starting with 1. Hisoffice is numbered 1, and his house is numbered 2. The first lineof each test case gives the number of intersections N, 1< N <= 1000, and the number of pathsM. The following M lines each contain a pair of intersections a band an integer distance 1 <= d <=1000000 indicating a path of length d between intersection a and adifferent intersection b. Jimmy may walk a path any direction hechooses. There is at most one path between any pair ofintersections.

Output

For each test case,output a single integer indicating the number of different routesthrough the forest. You may assume that this number does not exceed2147483647.

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

题意: 从A到B, 并且B到达终点最短路径小于A到终点的最短路径.要你求出一共存在多少条B开始的这样的路径.

 

解题思路:

   1.反过来求出终点到各个点的最短路. (正难则反)

   2.接着我们要求得时B到达终点并且小于A到终点的路径数目. 满足条件 u --> v, dis[v][t]> dis[i][t]

     (1 <= i <= n)

   3.采用当然是记忆化搜索.

 

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
#define MAX 1005
const int INF = 2147483647;

struct node
{
 int u,v;
 int w;
 int next;
}edges[MAX*2];

int n, m;
int first[MAX];
bool g[MAX][MAX], vis[MAX];
int dist[MAX];
int dp[MAX];
int num;

void addedge(int u,int v,int w)
{
 edges[num].u = u;
 edges[num].v = v;
 edges[num].w = w;
 edges[num].next = first[u];
 first[u] = num++;
}

void read_graph()
{
 memset(g,false,sizeof(g));
 memset(dp,-1,sizeof(dp));
 memset(first,-1,sizeof(first));
 num = 0;
 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);
  g[u][v] = g[v][u] = true;
 }
}

void spfa(int start)
{
 queue<int>qu;
 memset(vis,false,sizeof(vis));
 qu.push(start);
 vis[start] = true;
 int i;
 for(i = 1; i <= n; ++i) dist[i] =(i == start ? 0 : INF);

 while( !qu.empty() )
 {
  int u = qu.front();
  qu.pop();
  vis[u] = false;

  for(int e = first[u]; e !=-1; e = edges[e].next)
  {
   if(dist[edges[e].v]> dist[u] + edges[e].w)
   {
    dist[edges[e].v]= dist[u] + edges[e].w;
    if(!vis[edges[e].v] )
    {
     vis[edges[e].v]= true;
     qu.push(edges[e].v);
    }
   }
  }
 }
}

int solve(int v)
{
 if(dp[v] != -1) return dp[v];
 if(v == 2) return 1;
 dp[v] = 0;
 for(int i = 1; i <= n; ++i)
 {
  if(g[v][i]&& dist[v] >dist[i])
  {
   dp[v] +=solve(i);
  }
 }
 return dp[v];
}

int main()
{
// freopen("input.txt","r",stdin);
 while(scanf("%d",&n) != EOF&& n != 0)
 {
  scanf("%d",&m);
  read_graph();

  spfa(2);
  int result = solve(1);
  printf("%d\n",result);
 }

 return 0;
}