题目链接:https://vjudge.net/problem/UVA-10917 题目意思:Jimmy下班回家要闯过一下森林,劳累一天后在森林中散步是非常惬意的事,所以他打算每天沿着一条不同的路径回家,欣赏不同的风景,但他也不太想太晚回家,因此他不打算走回头路.换句话来说,他只会沿着如下条件的道路(A,B)走:存在一条从B出发回家的路径,比所有从A出发回家的路径都要短.我们的任务是要找出一共有有多少条不同的回家路径,家的编号是2,公司的编号是1. 题目思路:最短路+dp,我们现在考虑如何dp,…

题目大意:n个点,m条边的无向图.一个人从起点到终点按照下面的走法:从A走向B当A到终点的最小距离比B到终点的最小距离大时.问从起点到终点有多少路径方案. 题目分析:先用dijkstra预处理出终点到每个点的最短路,然后将满足行走条件的A.B(除行走条件外,还要满足一个前提,即A.B之间要有边)用一条有向边连起来(A->B),得到一个DAG,动态规划解决. 代码如下: # include<iostream> # include<cstdio> # include<vec…

A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4603    Accepted Submission(s): 1671 Problem DescriptionJimmy experiences a lot of stress at work these days, especially…

A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5679    Accepted Submission(s): 2086 Problem Description Jimmy experiences a lot of stress at work these days, especiall…

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…

A Walk Through the Forest Time Limit:1000MS  Memory Limit:65536K Total Submit:48 Accepted:15 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 t…

A Walk Through the Forest Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 3   Accepted Submission(s) : 1 Problem Description Jimmy experiences a lot of stress at work these days, especially since…

A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9081 Accepted Submission(s): 3353 Problem Description Jimmy experiences a lot of stress at work these days, especially sin…

A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 118 Accepted Submission(s): 57   Problem Description Jimmy experiences a lot of stress at work these days, especially sinc…

Problem    UVA - 10917 - Walk Through the Forest Time Limit: 3000 mSec 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…

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…

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 ot…

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…

A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7733    Accepted Submission(s): 2851 Problem Description Jimmy experiences a lot of stress at work these days, especiall…

B. A Walk Through the Forest Time Limit: 1000ms Memory Limit: 32768KB 64-bit integer IO format: %I64d      Java class name: Main   Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax af…

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…

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 ot…

题目给一张有向图,问从起点1到终点2沿着合法的路走有种走法,合法的路指从u到v的路,v到终点的距离严格小于u到终点的距离. 先SPFA预处理出所有合法的路,然后这些路肯定形成一个DAG,然后DP一下就OK了,d[u]表示u到终点2的方案数. #include<cstdio> #include<cstring> #include<queue> using namespace std; #define INF (1LL<<60) #define MAXN 111…

题目: gbn最近打算穿过一个森林,但是他比较傲娇,于是他决定只走一些特殊的道路,他打算只沿着满足如下条件的(A,B)道路走:存在一条从B出发回家的路,比所有从A出发回家的路径都短.你的任务是计算一共有多少条不同的回家路径.其中起点的编号为1,终点的编号为2. 分析: 先求出每个点到终点的距离,根据题目要求找d[u]>d[v]的路径(u,v)走,计算路径数即可. 代码如下: #include<cstdio> #include<cstdlib> #include<cstr…

求出家到其他点的最短路径,题目的条件变成了u->v不是回头路等价于d[u]>d[v]. 然后根据这个条件建DAG图,跑dp统计方案数,dp[u] = sum(dp[v]). #include<bits/stdc++.h> using namespace std; , maxm = ; struct Edge { int v,w,nxt; }; #define PB push_back vector<Edge> edges; vector<int> G[max…

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…

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…

用新模板阿姨了一天,换成原来的一遍就ac了= = 题意很重要..最关键的一句话是说:若走A->B这条边,必然是d[B]<d[A],d[]数组保存的是各点到终点的最短路. 所以先做dij,由d[B]<d[A]可知,所走的路径上各点的d[]值是由大到小的,即是一个DAG,从而决定用记忆化搜索查找总的路径数. #include<stdio.h> #include<string.h> #include<algorithm> using namespace st…

看到这道题,想起了我家旁边的山! 那是一座叫做洪山寨的山,据说由当年洪秀全的小妾居住于此而得名! 山上盛产野果(很美味)! 好久没有爬上去了! #include<stdio.h> #include<string.h> #define MAX 100000000 int map[1010][1010]; int res[1010]; int dist[1010]; bool visited[1010]; int n,m; void Dijkstra(int v)   { int i,…

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];…

题意: 给你一个图,找最短路.但是有个非一般的的条件:如果a,b之间有路,且你选择要走这条路,那么必须保证a到终点的所有路都小于b到终点的一条路.问满足这样的路径条数 有多少,噶呜~~题意是搜了解题报告才明白的Orz....英语渣~ 思路: 1.1为起点,2为终点,因为要走ab路时,必须保证那个条件,所以从终点开始使用单源最短路Dijkstra算法,得到每个点到终点的最短路,保存在dis[]数组中. 2.然后从起点开始深搜每条路,看看满足题意的路径有多少条. 3.这样搜索之后,dp[1]就是从起…

题意: 一个人从公司回家,他可以从A走到B如果从存在从B出发到家的一条路径的长度小于任何一条从A出发到家的路径的长度. 问这样的路径有多少条. 思路: 题意并不好理解,存在从B出发到家的一条路径的长度小于任何一条从A出发到家的路径的长度,从这个条件可以推出只要满足B到家的最短路小于从A到家的最短路,那么就是满足条件的. 所以从家开始求到各个点的最短路,然后从公司开始进行记忆化搜索求出路径的总条数. 如果两个点A和B满足d[A] > d[B],那么A到家的路径条数一定包括B到家的路径条数,临界条件…

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…