A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6935    Accepted Submission(s): 2548

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
 
题意是求从起点1到终点2的满足条件的路径条数,条件是该条路径上的所有边AB都要满足A到终点的最短路大于B到终点的最短路。
思路就是Dijkstra+记忆化搜索

/*
ID: LinKArftc
PROG: 1142.cpp
LANG: C++
*/ #include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll; const int maxn = ;
int n, m;
int mp[maxn][maxn];
int dis[maxn];
bool vis[maxn]; void dij(int s) {
for (int i = ; i <= n; i ++) dis[i] = mp[s][i];
memset(vis, , sizeof(vis));
vis[s] = true;
dis[s] = ;
for (int i = ; i <= n; i ++) {
int mi = inf;
int ii = s;//要赋值,因为下一行的for循环可能并不改变ii的值,所以可能会RE
for (int j = ; j <= n; j ++) {
if (!vis[j] && dis[j] < mi) {
mi = dis[j];
ii = j;
}
}
vis[ii] = true;
for (int j = ; j <= n; j ++) {
if (!vis[j] && dis[j] > dis[ii] + mp[ii][j]) {
dis[j] = dis[ii] + mp[ii][j];
}
}
}
} int cnt[maxn]; int dfs(int cur) {
if (cnt[cur]) return cnt[cur];
if (cur == ) return ;
int ret = ;
for (int i = ; i <= n; i ++) {
if (mp[cur][i] == inf || dis[i] >= dis[cur]) continue;
ret += dfs(i);
}
cnt[cur] = ret;
return ret;
} int main() {
//input;
int u, v, c;
while (~scanf("%d", &n) && n) {
scanf("%d", &m);
memset(mp, 0x3f, sizeof(mp));
for (int i = ; i <= m; i ++) {
scanf("%d %d %d", &u, &v, &c);
mp[u][v] = c;
mp[v][u] = c;
}
dij();
//for (int i = 1; i <= n; i ++) printf("dis[%d] = %d\n", i, dis[i]);
memset(cnt, , sizeof(cnt));
printf("%d\n", dfs());
} return ;
}

HDU1142 (Dijkstra+记忆化搜索)的更多相关文章

  1. luogu3953 [NOIp2017]逛公园 (tarjan+dijkstra+记忆化搜索)

    先跑一边dijkstra算出从1到i的最短距离dis[i] 然后建反向边 从n开始记忆化搜索,(p,k)表示1到p的距离=dis[p]+k的方案数 答案就是$\sum\limits_{i=0}^{k} ...

  2. HDU 1142 A Walk Through the Forest(Dijkstra+记忆化搜索)

    题意:看样子很多人都把这题目看错了,以为是求最短路的条数.真正的意思是:假设 A和B 是相连的,当前在 A 处, 如果 A 到终点的最短距离大于 B 到终点的最短距离,则可以从 A 通往 B 处,问满 ...

  3. Luogu 3953[NOIP2017] 逛公园 堆优化dijkstra + 记忆化搜索

    题解 首先肯定是要求出单源最短路的,我用了堆优化dijikstra ,复杂度 mlogm,值得拥有!(只不过我在定义优先队列时把greater 打成了 less调了好久 然后我们就求出了$i$到源点的 ...

  4. hdu-1142(记忆化搜索+dij)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1142 思路:1.不是求最短路径,而是求如果两个点A,B直接相连,且A到终点的距离大于B到终点的距离,求 ...

  5. hdu1142(dj+记忆化搜索)

    题意:给你n各点,m行关于这些点的联通关系,以及距离,求从1这个点到2这个点之间,下一个点到2这个点比当前点到2这个点的距离要小的路径的条数...... 思路:dj+记忆化搜索....... #inc ...

  6. hduoj----1142A Walk Through the Forest(记忆化搜索+最短路)

    A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Jav ...

  7. HDU-1428(记忆化搜索)

    Problem Description LL 最近沉迷于AC不能自拔,每天寝室.机房两点一线.由于长时间坐在电脑边,缺乏运动.他决定充分利用每次从寝室到机房的时间,在校园里散散步.整个HDU 校园呈方 ...

  8. UVA - 10917 - Walk Through the Forest(最短路+记忆化搜索)

    Problem    UVA - 10917 - Walk Through the Forest Time Limit: 3000 mSec Problem Description Jimmy exp ...

  9. HDU 1142 A Walk Through the Forest(最短路+记忆化搜索)

    A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Jav ...

随机推荐

  1. 第二十篇 sys模块

    修改环境变量 import sys sys.path.append() 但是,这种修复方式只是临时修改 如果要永久修改,就要电脑里配置环境变量. sys.argv:命令行参数List,第一个元素是程序 ...

  2. lintcode-95-验证二叉查找树

    95-验证二叉查找树 给定一个二叉树,判断它是否是合法的二叉查找树(BST) 一棵BST定义为: 节点的左子树中的值要严格小于该节点的值. 节点的右子树中的值要严格大于该节点的值. 左右子树也必须是二 ...

  3. STL中list的erase()方法

    http://www.cnblogs.com/gshlsh17/ rase()方法是删除iterator指定的节点  但是要注意的是在执行完此函数的时候iterator也被销毁了   这样的话关于it ...

  4. web相关基础知识4

      一.定位的盒子居中 Css可见性 overflow: hidden;   溢出隐藏   常用在超出盒子之后就隐藏 visibility: hidden;   隐藏元素    隐藏之后还占据原来的位 ...

  5. 修改虚拟机上Linux系统的IP地址

    然后再输入:ifconfig eth0 192.168.11.6 netmask 255.255.255.0   . 这样就可以把网卡eth0的IP地址修改为 192.168.11.6

  6. python3 urllib和requests模块

    urllib模块是python自带的,直接调用就好,用法如下: 1 #处理get请求,不传data,则为get请求 2 import urllib 3 from urllib.request impo ...

  7. BZOJ4424/CF19E Fairy(dfs树+树上差分)

    即删除一条边使图中不存在奇环.如果本身就是个二分图当然任意一条边都可以,先check一下.否则肯定要删除在所有奇环的交上的边. 考虑怎么找这些边.跑一遍dfs造出dfs树,找出返祖边构成的奇环.可以通 ...

  8. Failed with exception MetaException(message:javax.jdo.JDODataStoreException: Error(s) were found while auto-creating/validating the datastore for classes.

    hive (db_emp)> load data local inpath '/opt/datas/emp.txt' into table emp_part partition(`date`=' ...

  9. NOIP2010 codevs1069 洛谷P1525 关押罪犯

    Problem Description S 城现有两座监狱,一共关押着N 名罪犯,编号分别为1~N.他们之间的关系自然也极 不和谐.很多罪犯之间甚至积怨已久,如果客观条件具备则随时可能爆发冲突.我们用 ...

  10. 【NOIP模拟赛】公主的朋友 区间染色问题

    这道题大家都用的分块,然而我发现这是一个经典算法:区间染色问题. 我们区间染色时把区间分成若干连续的颜色段,然后我们每次染色删除原来的颜色段插入新的颜色段. 我们发现我们的时间复杂度直接与我们要染色区 ...