标签：href template fine eof pos 上下 code htm map
题目链接点我
题目大意：给你一个nxm行的01矩阵，你每次可以点一个位置然后反转这个位置上下左右以及自己的状态，问你是否可以把它全置为0，如果可以，把字典序最小的情况输出出来。
??这题和我上一篇博文类似传送门←_←不同的是，题目要把变换次数最小的情况下字典序最小的结果输出出来。其实字典序不用担心，因为我们二进制枚举本来就是按字典序从低到高来的，
所以我们只要关心次数就行了。
//https://www.cnblogs.com/shuitiangong/
#include<set>
#include<map>
#include<list>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define endl '\n'
#define rtl rt<<1
#define rtr rt<<1|1
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define maxx(a, b) (a > b ? a : b)
#define minn(a, b) (a < b ? a : b)
#define zero(a) memset(a, 0, sizeof(a))
#define INF(a) memset(a, 0x3f, sizeof(a))
#define IOS ios::sync_with_stdio(false)
#define _test printf("==================================================\n")
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<ll, ll> P2;
const double pi = acos(-1.0);
const double eps = 1e-7;
const ll MOD = 1000000007LL;
const int INF = 0x3f3f3f3f;
const int _NAN = -0x3f3f3f3f;
const double EULC = 0.5772156649015328;
const int NIL = -1;
template<typename T> void read(T &x){
x = 0;char ch = getchar();ll f = 1;
while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
const int maxn = 17;
int dx[] = {0, 0, 0, 1, -1};
int dy[] = {1, -1, 0, 0, 0};
int n, m, cnt;
char g[maxn][maxn], bk[maxn][maxn], ans[maxn][maxn], tmp[maxn][maxn];
void turn(int x, int y) { //改变自己&&周围4个点的状态
++cnt;
tmp[x][y] = '1';
for (int i = 0; i<5; ++i) {
int xx = x+dx[i], yy = y+dy[i];
if (xx>=0 && xx<n && yy>=0 && yy<m)
bk[xx][yy] ^= 1;
}
}
int main(void) {
while(~scanf("%d%d%*c", &n, &m)) {
for (int i = 0; i<n; ++i)
for (int j = 0; j<m; ++j)
scanf("%c%*c", &g[i][j]);
int __max = INF;
for (int k = 0; k<1<<m; ++k) {
cnt = 0;
memcpy(bk, g, sizeof(g)); //复制初始状态
for (int i = 0; i<n; ++i) //默认每个点的点击数目全为0
for (int j = 0; j<m; ++j)
tmp[i][j] = '0';
for (int i = 0; i<m; ++i) //二进制枚举第一行所有情况
if (k>>i&1) turn(0, i);
for (int i = 0; i<n-1; ++i) //根据上一行状态递推下一行操作
for (int j = 0; j<m; ++j)
if (bk[i][j]&1) turn (i+1, j);
bool flag = true;
for (int i = 0; i<m; ++i) //检查最后一行是否全为0
if (bk[n-1][i]&1) flag = false;
if (flag && __max > cnt) {
__max = cnt;
memcpy(ans, tmp, sizeof(tmp));
}
}
if (__max != INF)
for (int i = 0; i<n; ++i)
for (int j = 0; j<m; ++j)
printf(j==m-1 ? "%c\n" : "%c ", ans[i][j]);
else printf("IMPOSSIBLE\n");
}
return 0;
}标签：href template fine eof pos 上下 code htm map
原文地址：https://www.cnblogs.com/shuitiangong/p/12518927.html
