POJ 3279 Fliptile

最新推荐文章于 2021-07-06 00:06:22 发布

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最新推荐文章于 2021-07-06 00:06:22 发布

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本文探讨了一种关于Fliptile问题的解决策略，即如何通过最小化翻转次数来完成特定任务，并在多个可能的解决方案中选择字典序最小的一个。通过枚举第一排的翻转情况，结合后续排的翻转策略，最终确定了可行的最优方案。此问题涉及到贪心算法与状态转移方程的应用。

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Fliptile

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5719		Accepted: 2162

Description

Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N≤ 15) of square tiles, each of which is colored black on one side and white on the other side.

As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make.

Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE".

Input

Line 1: Two space-separated integers: M and N
Lines 2.. M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white

Output

Lines 1.. M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.

Sample Input

Sample Output

Source

USACO 2007 Open Silver

kuangbin带你飞专题1 D题

感觉这个题有点神。。。

题意就是给你m*n的矩阵有0 有1 需要把所有1翻转成0

每次选i,j翻转会同时翻转i-1,j i+1,j i,j-1 i,j+1

问次数最少的方案

如果有多种输出字典序最小的

首先每个位置最多翻一次，因为翻两次和没翻效果是一样的

但是想把第一排全变成0 可以由第一排翻转也可以由第二排翻转所以不太好做

于是可以枚举第一排的反转情况第一排翻转完成后想让第一排全变成0 只能通过翻转第二排的某几个

同时第二排也确定了只能翻转第三排

就可以推出来所有的应该翻转的情况了

然后最后一排反转情况确定了最后的翻转完成情况就确定了如果最后一排全是0 那么是一个可行方案否则不可行

按字典序枚举第一排翻转情况能保证答案是按字典序排列的

然后记录翻转次数

次数小于时更新答案

金大爷太神辣！

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int mx[5]={0,1,0,-1,0};
const int my[5]={0,0,1,0,-1};
int f[22][22];
int m,n;
int cz[22][22];
int ok;
int t[22][22];
int ans;
int p[22][22];
void work(int x,int y)
{
    for(int i=0;i<=4;i++)
    {
        int xx=x+mx[i],yy=y+my[i];
        t[xx][yy]^=1;
    }
}

int test()
{
    int ret=0;
    for(int i=1;i<=n;i++)
        if(cz[1][i])
            ret++;
            
    for(int i=2;i<=m;i++)
        for(int j=1;j<=n;j++)
            if(t[i-1][j]==1)
            {
                cz[i][j]=1;
                work(i,j);
                ret++;
            }
            else
                cz[i][j]=0;
    for(int i=1;i<=n;i++)
        if(t[m][i]!=0)
            return -1;
    return ret;
}
void dfs(int dep)
{
    if(dep==n+1)
    {
        memcpy(t,f,sizeof(f));
        for(int i=1;i<=n;i++)
            if(cz[1][i])
                work(1,i);
        int k=test();
        if(k!=-1)
        {
            if(k<ans)
            {
                ans=k;
                memcpy(p,cz,sizeof(cz));
            }
        }
        return;
    }
    for(int i=0;i<=1;i++)
    {
        cz[1][dep]=i;
        dfs(dep+1);
    }
}
int main()
{
    while(scanf("%d%d",&m,&n)==2)
    {
        ans=0x3f3f3f3f;
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&f[i][j]);
        dfs(1);
        if(ans!=0x3f3f3f3f)
        {
            for(int i=1;i<=m;i++)
            {
                for(int j=1;j<=n;j++)
                    printf("%d ",p[i][j]);
                printf("\n");
            }
        }
        else
        {
            printf("IMPOSSIBLE\n");
        }
    }
    return 0;
}