UVA 10318 Security Panel（DFS剪枝 + 状压 + 思维）题解

题意：给一个r*c的矩阵开关（初始全打开的），每次按下一个开关都会改变3*3范围内的有*的地方的状态，问你最少几步能让开关全闭上，按升序输出按哪些按钮

思路：每个按钮至多按一下，按按钮的顺序和结果无关。我们把当前矩阵的开关状态状压到 long long，并且预处理按下每个按钮的变化，这样可以直接异或得到新的状态。这里还需要剪枝，因为顺序无关，我们直接从1按到r*c，那么如果我们按到第k行，现在就已经影响不到k-2行了，所以k-2行如果有开关没闭上就return。

代码：

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 1e5 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int r, c;
char s[10][10];
ll change[30], ansStep[30], step[30], vis[10][10], lit, ansCnt, cnt, OK;
void init(int x, int y){
    int pos = (x - 1) * c + y;
    change[pos] = 0;
    for(int i = 1; i <= 3; i++){
        for(int j = 1; j <= 3; j++){
            if(s[i][j] == '.') continue;
            int xx = x + i - 2, yy = y + j - 2;
            if(xx < 1 || xx > r || yy < 1 || yy > c) continue;
            ll p = (xx - 1) * c + yy;
            change[pos] += (1LL << p);
        }
    }
}
void dfs(int pos){
    if(pos > r * c) return;
    int x = pos / c, y = pos % c;
    if(x >= 3){
        for(int i = 1; i <= c; i++){
            int pos2 = (x - 3) * c + i;
            if(!(lit & (1LL << pos2))) return;
        }
    }
    lit ^= change[pos];
    vis[x][y] = 1;
    step[cnt++] = pos;
    if(lit == OK){
        if(cnt < ansCnt){
            for(int i = 0; i < cnt; i++)
                ansStep[i] = step[i];
            ansCnt = cnt;
            lit ^= change[pos];
            vis[x][y] = 0;
            cnt--;
            return;
        }
    }
    dfs(pos + 1);
    lit ^= change[pos];
    vis[x][y] = 0;
    cnt--;
    dfs(pos + 1);
}
int main(){
    int ca = 1;
    while(~scanf("%d%d", &r, &c) && r + c){
        OK = 0;
        for(int i = 1; i <= 3; i++) scanf("%s", s[i] + 1);
        for(int i = 1; i <= r; i++){
            for(int j = 1; j <= c; j++){
                init(i, j);
                OK += (1LL << ((i - 1) * c + j));
            }
        }
        ansCnt = 100;
        memset(vis, 0, sizeof(vis));
        dfs(1);
        printf("Case #%d
", ca++);
        if(ansCnt == 100) printf("Impossible.
");
        else{
            for(int i = 0; i < ansCnt; i++){
                if(i != 0) printf(" ");
                printf("%d", ansStep[i]);
            }
            printf("
");
        }
    }
    return 0;
}