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,问满足这种的条件的从办公室到家的路径条数. 分析:1.以终点 2 为起点 Dijkstra跑一边最短路,找到所有点到2的最短距离:        2.直接DFS记忆化搜索. 注意:记忆化搜索时的return值,否则此很容易TLE 解法1:O(n^2) #include<iostream> #inclu…
A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4383    Accepted Submission(s): 1573 Problem Description Jimmy experiences a lot of stress at work these days, especiall…
A Walk Through the Forest 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1142 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. T…
http://acm.hdu.edu.cn/showproblem.php?pid=1142 这道题是spfa求最短路,然后dfs()求路径数. #include <cstdio> #include <queue> #include <cstring> #include <algorithm> #define maxn 1001 using namespace std; <<; int g[maxn][maxn]; int dis[maxn];…
A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10172    Accepted Submission(s): 3701 Problem Description Jimmy experiences a lot of stress at work these days, especial…
A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5306    Accepted Submission(s): 1939 Problem Description Jimmy experiences a lot of stress at work these days, especiall…
题目链接:acm.hdu.edu.cn/showproblem.php?pid=1142 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,…
A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8850    Accepted Submission(s): 3267 Problem Description Jimmy experiences a lot of stress at work these days, especiall…
原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1142 题目大意:Jimmy要从办公室走路回家,办公室在森林的一侧,家在另一侧,他每天要采取不一样的路线回家.由于他要尽快回家,他在选择路线的时候总是要越来越靠近他家.计算符合条件的路线一共有几种. 解题思路:题目要求“路线要越来越靠近家”,也就是说每次选择下一个结点的时候距离家的距离比当前的结点近.首先该结点离家的距离就是该节点到家的最短路径的长度.所以我们先求出所有节点到家的最短路径.(Dijks…
题目链接 题意 :办公室编号为1,家编号为2,问从办公室到家有多少条路径,当然路径要短,从A走到B的条件是,A到家比B到家要远,所以可以从A走向B . 思路 : 先以终点为起点求最短路,然后记忆化搜索. #include <cstdio> #include <queue> #include <cstring> #include <iostream> << ; using namespace std ; int N,M ; ][] ,pre[],d…