https://blog.csdn.net/guhaiteng/article/details/52730373 参考题解 http://codeforces.com/contest/723/problem/D  原题目 #include<iostream> #include<cstdio> #include <cctype> #include<algorithm> #include<cstring> #include<cmath>…
Oil Deposits Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64 Description The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of…
D. Lakes in Berland time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The map of Berland is a rectangle of the size n × m, which consists of cells of size 1 × 1. Each cell is either land or…
D. Lakes in Berland time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The map of Berland is a rectangle of the size n × m, which consists of cells of size 1 × 1. Each cell is either land or…
题意:给定一个n*m的矩阵,*表示陆地, . 表示水,一些连通的水且不在边界表示湖,让你填最少的陆地使得图中湖剩下恰好为k. 析:很简单的一个搜索题,搜两次,第一次把每个湖的位置和连通块的数量记下来,第二次去填陆地,选少的进行填. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdli…
题目链接: C. Three States time limit per test 5 seconds memory limit per test 512 megabytes input standard input output standard output The famous global economic crisis is approaching rapidly, so the states of Berman, Berance and Bertaly formed an allia…
#include "iostream" #include "cstdio" using namespace std; ][]={{,},{,-},{,},{-,},{,-},{-,},{,},{-,-}}; int count,r,c; ][]; void dfs(int x,int y)//深搜函数,参数为坐标(定位) { ||y<||x>=r||y>=c) return ; else if(map[x][y]=='@') { map[x][y]…
Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer…
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5268 题目大意:字符一样并且相邻的即为连通.每次可翻转一个连通块X(O)的颜色,问至少改变几次使得图上所有字符都相等. 解题思路: 1) dfs( 建图 ) ,因为翻转的时候每翻转连通块中一个整个连通块都翻转,这样你可以将其看成一个有边相连的无向图,每个边的两个顶点颜色都不一样. 2) bfs( 寻找最优解 ) , 建完图后就需要翻转计算最优解,可以枚举从每一点开始…
A /*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; ][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }}; ; + + + + 1e9…