题目链接:pid=4462">传送门 题意:一个n*n的区域,有m个位置是能够放稻草人的.其余都是玉米.对于每一个位置(x,y)所放稻草人都有个作用范围ri, 即abs(x-i)+abs(y-j)<=r,(i,j)为作用范围内.问至少要在几个位置上放稻草人,才干覆盖全部的玉米,若不可能则输出-1. 有一个trick,就是放稻草人的位置不用被覆盖 eg: input: 2 4 1 1 1 2 2 1 2 2 0 0 0 0 output: 0 0 代码例如以下: #include &l…

又一发吐血ac,,,再次明白了用函数(代码重用)和思路清晰的重要性. 11779687 2014-10-02 20:57:53 Accepted 4770 0MS 496K 2976 B G++ czy Lights Against Dudely Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1360    Accepted Subm…

题意: N*N的矩阵中有M个点能够放稻草人.且给覆盖距离R 每一个稻草人能覆曼哈顿距离R以内的点 问最少须要多少个稻草人 思路: 由于范围非常小,直接能够暴力 注意稻草人所在的位置是不须要被覆盖的 代码: #include"cstdlib" #include"cstdio" #include"cstring" #include"cmath" #include"queue" #include"alg…

题意:给定一个 n*n的矩阵,在一些位置放上稻草人,每个稻草人的范围是一定,问你最少几个能覆盖整个矩阵. 析:稻草人最多才10个,所以考虑暴力,然后利用二进制法,很容易求解,并且时间很少0ms,注意有一个坑,就是那些指定的位置是可以不用覆盖的, 当时WA一次. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream&…

It's harvest season now! Farmer John plants a lot of corn. There are many birds living around his corn field. These birds keep stealing his corn all the time. John can't stand with that any more. He decides to put some scarecrows in the field to driv…

题目 这是一道可以暴力枚举的水题. //以下两个都可以ac,其实差不多一样,呵呵 //1: //4 wei shu #include<stdio.h> struct tt { ],b[],c[]; }e[]; int main() { ],mark[],yi,flag,a1,a2,a3,a4; while(scanf("%d",&n),n) { ;i<n;i++) { scanf("%s%s%s",e[i].a,e[i].b,e[i].c)…

Crazy Tank Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4562    Accepted Submission(s): 902 Problem Description Crazy Tank was a famous game about ten years ago. Every child liked it. Time f…

HDU 4930 Fighting the Landlords 题目链接 题意:就是题中那几种牌型.假设先手能一步走完.或者一步让后手无法管上,就赢 思路:先枚举出两个人全部可能的牌型的最大值.然后再去推断就可以 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; struct Player { int rank[15]; } p1, p2; int t,…

题意:... 析:我们可以知道,a1+a2=b1,那么我们可以枚举a1,那么a2就有了,并且a1+a3=b2,所以a3就有了,我们再从把里面的剩下的数两两相加,并从b数组中去掉, 那么剩下的最小的就是a4,然后依次可以求出a5,a6....由于a最大才是5000,并且保证有唯一解,那么找到一个就直接退出. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #incl…

Problem Description Let L denote the number of 1s in integer D’s binary representation. Given two integers S1 and S2, we call D a WYH number if S1≤L≤S2. With a given D, we would like to find the next WYH number Y, which is JUST larger than D. In othe…