hdu 3935 Dark Room 有n*m的灯，最少几次可以把所有的灯点亮,每次可点周围5个 m<=20

解决灯光控制问题：最小点亮次数算法

最新推荐文章于 2020-05-16 11:24:50 发布

原创最新推荐文章于 2020-05-16 11:24:50 发布 · 1.4k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#output #wizard #integer #input #matrix #each

acm_数学问题专栏收录该内容

163 篇文章

订阅专栏

本文探讨了一个在魔法房间中通过控制特定位置的灯泡，最小化点亮所有灯泡所需操作次数的问题。利用分治思想和二进制操作，文章详细介绍了算法实现过程及其实现代码。

Problem Description

One day Harry Potter came to a magic room, in which there are some shadow monsters sent by the black wizard waiting for him. Fortunately, shadow monsters can only live in areas where there are no lights shining, there are n*m lamps on the roof of the room, every little lamp can only illuminate the scope of the fixed area, and all the area of lights shining don't overlap, some lamps on, some lamps off, the areas below the closed lamps will appear the shadow, and in these areas shadow monsters will attack Harry Potter. Now Harry Potter comes to you and gives you the previous state of the room lights, Harry Potter can control the light switch in a remote place .The state of a light will change as follows: if a lamp state changed, the state of the lights which before it, after it, left to it or right to it will all change. Your task is to calculate how many times we need at least to turn on all the lights by changing the state of the lamps.

Input

in the first line there are two integers, n (1 < = n < = 100) and m (1 < = m < = 15), stands for there are n lines, each line have m data, 0 stands for the lights off, 1 stands for the lights on.

Output

for each group of input data, output an integer stands for the minimum number we need to change the state of the lights to put all the lights on, if we can't turn on all the lamps, output 'no solution'.

Sample Input

Sample Output

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=100000;
const int inf=(1<<28);
//有n*m的灯，最少几次可以把所有的灯点亮,每次可点周围5个 m<=20
int g[maxn],x[maxn];
int n,m;
int all;
int ans;
//计算n二进制中1的个数分治思想 Matrix67博客有讲
int getOnesInN(int n)
{
    n=(n&0x55555555)+((n>>1)&0x55555555);
    n=(n&0x33333333)+((n>>2)&0x33333333);
    n=(n&0x0f0f0f0f)+((n>>4)&0x0f0f0f0f);
    n=(n&0x00ff00ff)+((n>>8)&0x00ff00ff);
    n=(n&0x0000ffff)+((n>>16)&0x0000ffff);
    return n;
}
int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        memset(g,0,sizeof(g));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                //0暗 1亮
                int tmp;scanf("%d",&tmp);
                g[i]=(g[i]<<1)|tmp;
            }
        }
        all=(1<<m)-1;
        ans=inf;
        for(int i=0;i<=all;i++)//枚举第一行开灯情况1,change;0,stay
        {
            for(int j=0;j<n;j++) x[j]=g[j];//用memcpy会超时
            int cnt=getOnesInN(i);//开灯次数
            //press 0 i
            x[0]^=i,x[0]^=i<<1,x[0]^=i>>1;x[0]&=all;
            x[1]^=i;
            //2-n行已经确定
            for(int j=1;j<n;j++)
            {
                int k=x[j-1]^all;//上一行0的位置
                //press j,k
                x[j]^=k,x[j]^=k<<1,x[j]^=k>>1;x[j]&=all;
                x[j+1]^=k;
                //
                cnt+=getOnesInN(k);
            }
            if(x[n-1]==all) ans=min(ans,cnt);
        }
        if(ans==inf)printf("no solution\n");
        else printf("%d\n",ans);
    }
    return 0;
}