Free DIY Tour Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2735    Accepted Submission(s): 889 [题目链接]http://acm.hdu.edu.cn/showproblem.php?pid=1224 [题目意思]题目是说给你一个有环的图,结点之间有路径,每个结点都有个兴趣值,现在给你起…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1224 Free DIY Tour Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8692    Accepted Submission(s): 2804 Problem Description Weiwei is a software en…

   Free DIY Tour Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6882    Accepted Submission(s): 2241    Problem Description Weiwei is a software engineer of ShiningSoft. He has just excellentl…

路径只能由小序号到大序号..(起点可以视为最小的序号和最大的序号) 问怎么走 happy值最大.. DFS N=100 且只能小序号到大序号 显然dfs可以过.. 但是存路径的时候sb了.....应该在更新答案时拿一个ans_dis数组去全部存一遍路径... #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <ctime> #incl…

一道很好的dfs加储存路径的题目  :路径保存:每次dfs都存i 当大于max时 将临时数组保存到答案数组 并不是当 当前值大于最大值时更新路径 还要加上一个条件:能回去 #include<bits/stdc++.h> using namespace std; int n; ][]; ]; ];]; int maxi,len; void dfs(int stepn,int sum,int cur) { ;i<=n;i++) { if(m1[cur][i]) { path[stepn]=i…

传送门 题目大意: 一个有向图(n + 1相当于1),每个点有一个权值(可以认为1和n+1权值为0),求从1走到n+1(相当于走回1)的最大路径权值和是多少,输出方案. 题目分析: 最短路问题,输出方案只需在dijkstra更新时记录from数组,完成后倒推即可. code #include<bits/stdc++.h> using namespace std; typedef long long ll; namespace IO{ inline ll read(){ ll i = 0, f…

这道题可用动态规划也可以用搜索,下面都写一下 Description Weiwei is a software engineer of ShiningSoft. He has just excellently fulfilled a software project with his fellow workers. His boss is so satisfied with their job that he decide to provide them a free tour around t…

本题的背景是求定点和定点之间的最短路问题(所有的最短路 不是一个解  是全部解,方法手段来自数据结构课程中的迪杰斯特拉算法和dfs(深度优先遍历). 分别用两种方法编程如下代码 dfs #include <iostream> #include <cstring> #include <cstdio> using namespace std; #define maxv 510 bool visit[maxv]; int arc[maxv][maxv];//邻接矩阵 int…

Weiwei is a software engineer of ShiningSoft. He has just excellently fulfilled a software project with his fellow workers. His boss is so satisfied with their job that he decide to provide them a free tour around the world. It's a good chance to rel…

题目大意:每一个城市都有一定的魅力值,然后有一个有向图,根据这个有向图从1到n+1所获得的魅力的最大值,并输出路径(要求只能从编号娇小的城市到编号较大的城市). 题解:很容易想到最短路+路径纪录.但是感觉有点小题大做了.我开始的方法是dfs+dp,dp[i]表示i的子节点最大的魅力值,但是它给的是一个图,并不是树,其中有环,所以之一WA...... 正解:两个for循环,dp[i]表示从第1个城市到第i个城市的最好状态.然后枚举小于i的所有城市, 状态转移方程dp[i]=max(dp[i],dp…