hdu 1142 A Walk Through the Forest（最短路+记忆化搜索dfs）

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2183    Accepted Submission(s): 787

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
  

 

             题目大意：寻找一共有多少条符合题意的路。能够从点A走到点B的要求是：点A到终点的最短路 > 点B到终点的最短路。 也就是说：从终点出发，求每一个点的最短路，然后那些最短路的值记录起来，作为能否通过的判断条件。最后用记忆化搜索来搜索出一共多少条符合要求的路。普通的dfs是超时的，bfs是超内存的。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1142

代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <queue>
#include <vector>
using namespace std;

const int INF = 99999999;
const int N = 1005;

vector<int> road[N];    //定义一个向量（第一次用-O(∩_∩)O-）
int map[N][N], dist[N], s[N];
bool visit[N];
int n, t, sum;

void init()
{
    int i, j;
    sum = 0;
    memset(s, 0, sizeof(s));
    for(i = 0; i <= n; i++) road[i].clear();
    for(i = 0; i <= n; i++)
        for(j = 0; j <= n; j++)
            if(i == j) map[i][j] = 0;
            else map[i][j] = INF;
}

void input()
{
    int a, b, d;
    while(t--)
    {
        scanf("%d %d %d", &a, &b, &d);
        if(d < map[a][b])
        {
            map[a][b] = map[b][a] = d;  //双向的
            road[a].push_back(b);
            road[b].push_back(a);
        }
    }
}

void spfa()     //比较喜欢用spfa
{
    int i, now;
    memset(visit, false, sizeof(visit));
    for(i = 1; i <= n; i++) dist[i] = INF;
    dist[2] = 0;
    queue<int> Q;
    Q.push(2);
    visit[2] = true;
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();
        visit[now] = false;
        for(i = 1; i <= n; i++)
        {
            if(dist[i] > dist[now] + map[now][i])
            {
                dist[i] = dist[now] + map[now][i];
                if(visit[i] == false)
                {
                    Q.push(i);
                    visit[i] = true;
                }
            }
        }
    }
}

int dfs(int now)    //记忆化搜索
{
    int i;
    if(now == 2) return 1;  //找到终点，返回1（1条路）
    if(s[now]) return s[now];   //该点已经走过，返回：过该点有多少条路
    for(i = 0; i < road[now].size(); i++)
    {
        if(dist[now] > dist[ road[now][i] ])
        {
            s[now] += dfs(road[now][i]);
        }
    }
    return s[now];  //返回：该点一共可以有多少条路
}

int main()
{
    while(scanf("%d", &n), n)
    {
        scanf("%d", &t);
        init();
        input();
        spfa();
        sum = dfs(1);   //sum一共多少条路，sum = s[1] 也行
        printf("%d\n", sum);
    }

    return 0;
}