题意:给n个点,每个点有一个人,有n-1条有权值的边,求所有人不在原来位置所移动的距离的和最大值. 析:对于每边条,我们可以这么考虑,它的左右两边的点数最少的就是要加的数目,因为最好的情况就是左边到右边,右边到左边,然后用dfs就可以解决了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <…

Holiday's Accommodation Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 200000/200000 K (Java/Others)Total Submission(s): 2009    Accepted Submission(s): 558 Problem Description Nowadays, people have many ways to save money on accommodation w…

Holiday's Accommodation Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 200000/200000 K (Java/Others)Total Submission(s): 2925    Accepted Submission(s): 894 Problem Description Nowadays, people have many ways to save money on accommodation w…

Problem Description Nowadays, people have many ways to save money on accommodation when they are on vacation. One of these ways is exchanging houses with other people. Here is a group of N people who want to travel around the world. They live in diff…

题目链接:  HDU 4118 Holiday's Accommodation 分析: 可以知道每条边要走的次数刚好的是这条边两端的点数的最小值的两倍. 代码: #include<iostream> #include<cstdio> #include<cstring> #include<stack> using namespace std; const int maxn=100000+10; struct node{ int to, dix, next; }…

题目链接: Holiday's Accommodation Time Limit: 8000/4000 MS (Java/Others)     Memory Limit: 200000/200000 K (Java/Others) Problem Description   Nowadays, people have many ways to save money on accommodation when they are on vacation.One of these ways is e…

HDU 1241 题目大意:给定一块油田,求其连通块的数目.上下左右斜对角相邻的@属于同一个连通块. 解题思路:对每一个@进行dfs遍历并标记访问状态,一次dfs可以访问一个连通块,最后统计数量. /* HDU 1241 Oil Deposits --- 入门DFS */ #include <cstdio> int m, n; //n行m列 ][]; /* 将和i,j同处于一个连通块的字符标记出来 */ void dfs(int i, int j){ || j < || i >=…

HDU 1241  Oil Deposits L -DFS Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u   Description The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large r…

HDOJ(HDU).1258 Sum It Up (DFS) [从零开始DFS(6)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (DF…

HDOJ(HDU).1016 Prime Ring Problem (DFS) [从零开始DFS(3)] 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (…