Fliptile

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10457		Accepted: 3866

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".

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

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

4 4
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1

0 0 0 0
1 0 0 1
1 0 0 1
0 0 0 0

USACO 2007 Open Silver

 

 

只要枚举第一行的格子的翻转情况，然后下面的行数翻转使得上面的格子变为白色，枚举完之后前(n-1)行一定全白，只要检查最后一行就ok

看的题解的思路自己没想到这样写哎

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int N,M;
int e1[20][20];
int e2[20][20];
int tim[20][20];
int an[20][20];
bool e3[20];
int ans=inf;
int fx[4][2]={1,0,-1,0,0,1,0,-1};
void turn(int x,int y)
{
    e2[x][y]=(e2[x][y]==0?1:0);
    for(int i=0;i<4;++i){
        int dx=x+fx[i][0];
        int dy=y+fx[i][1];
         if(dx<1||dy<1||dx>N||dy>M) continue;
                 e2[dx][dy]=(e2[dx][dy]==0?1:0);
    }
}
void make()
{
  int i,j;
  for(i=2;i<=N;++i){
    for(j=1;j<=M;++j){
        int up=e2[i-1][j];
        if(up==0) continue;
        else{
           turn(i,j);
           tim[i][j]++;
        }
    }
  }int s=0;
  for(i=1;i<=M;++i) if(e2[N][i]==0) s++;

  if(s==M){
    s=0;

    for(i=1;i<=N;++i){
        for(j=1;j<=M;++j) {
                s+=tim[i][j];
       // cout<<tim[i][j];
       // if(j!=M) cout<<" ";
        }
        //cout<<endl;
    }

    if(s<ans){
        ans=s;
        memcpy(an,tim,sizeof(int)*400);
    }
  }
}
void solve()
{
   int i,j,k;
   memcpy(e2,e1,sizeof(int)*400);
   memset(tim,0,sizeof(tim));
   for(i=1;i<=M;++i){
    if(e3[i]){
        turn(1,i);
        tim[1][i]++;
    }
   }
   make();
}
void dfs(int x)
{
  if(x==M+1){
    solve();
    return;
  }
  for(int i=0;i<2;++i){
    e3[x]=i;
    dfs(x+1);
  }
}
int main()
{
    int i,j,k;
    while(cin>>N>>M){ans=inf;
        for(i=1;i<=N;++i)
            for(j=1;j<=M;++j) {
                    cin>>e1[i][j];
            }
        dfs(1);
        if(ans==inf) puts("IMPOSSIBLE");
        else{
            for(i=1;i<=N;++i){
                for(j=1;j<=M;++j){
                    if(j!=1) cout<<" ";
                    cout<<an[i][j];
                }
                cout<<endl;
            }
        }
    }
    return 0;
}