【BZOJ】1647: [Usaco2007 Open]Fliptile 翻格子游戏(暴力)
http://www.lydsy.com/JudgeOnline/problem.php?id=1647
自己太弱。。。看题解。。
竟然是枚举第一行的放法,,,因为一定要全部变0,所以将前一行1的在这一行翻转掉就好了。。。因为是一定要翻掉前一行,所以正确性显然。。。。。。。。。。。。
T_T
表示智商不够
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; } const int N=20, dx[]={-1, 1, 0, 0}, dy[]={0, 0, -1, 1};
int c[N][N], ans[N][N], n, m, tot, mn=~0u>>1;
bool a[N][N], b[N][N]; void rot(int x, int y) {
b[x][y]=!b[x][y]; ++c[x][y];
rep(i, 4) {
int fx=dx[i]+x, fy=dy[i]+y;
if(fx<1 || fy<1 || fx>n || fy>m) continue;
b[fx][fy]=!b[fx][fy];
}
}
void getans(int x) {
CC(c, 0); tot=0;
for1(i, 1, n) for1(j, 1, m) b[i][j]=a[i][j];
for1(i, 0, m-1) if((1<<i)&x) { rot(1, i+1); ++tot; if(tot>=mn) return; }
for1(i, 2, n) for1(j, 1, m) if(b[i-1][j]) { rot(i, j); ++tot; if(tot>=mn) return; }
for1(i, 1, n) for1(j, 1, m) if(b[i][j]) return;
mn=tot;
for1(i, 1, n) for1(j, 1, m) ans[i][j]=c[i][j];
} int main() {
read(n); read(m);
for1(i, 1, n) for1(j, 1, m) read(a[i][j]);
int end=(1<<m)-1;
for1(i, 0, end) getans(i);
if(mn!=~0u>>1) for1(i, 1, n) { printf("%d", ans[i][1]); for1(j, 2, m) printf(" %d", ans[i][j]); puts(""); }
else puts("IMPOSSIBLE");
return 0;
}
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 x 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
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1
Sample Output
1 0 0 1
1 0 0 1
0 0 0 0
OUTPUT DETAILS:
After flipping at row 2 column 1, the board will look like:
0 0 0 1
1 0 1 0
1 1 1 0
1 0 0 1
After flipping at row 2 column 4, the board will look like:
0 0 0 0
1 0 0 1
1 1 1 1
1 0 0 1
After flipping at row 3 column 1, the board will look like:
0 0 0 0
0 0 0 1
0 0 1 1
0 0 0 1
After flipping at row 3 column 4, the board will look like:
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
Another solution might be:
0 1 1 0
0 0 0 0
0 0 0 0
0 1 1 0
but this solution is lexicographically higher than the solution above.
HINT
Source
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