题目链接

(bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=2669
(luogu) https://www.luogu.org/problem/P3160

题解

这道题充分暴露了我的菜。。

显然两个局部极小值点不能相邻，所以最多有\(8\)个局部极小值。
然后考虑容斥掉.不能成为局部极小值的限制，那么就变成钦定某些位置一定是局部极小值，其余不限，求方案数。

然后这个可以状压DP，考虑从小到大加入每个数，然后就很好求了。

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iostream>
#include<vector>
using namespace std;const int P = 12345678;
const int dx[8] = {1,0,-1,0,1,1,-1,-1},dy[8] = {0,1,0,-1,1,-1,-1,1};
vector<int> kx,ky;
int bitcnt[(1<<8)+3];
char a[7][11];
char b[7][11];
int num[(1<<8)+3];
int dp[31][(1<<8)+3];
int n,m,cnt,ans;bool check(int x,int y,int typ)
{bool ret = true;for(int i=0; i<8; i++){int xx = x+dx[i],yy = y+dy[i];if(xx>0&&xx<=n&&yy>0&&yy<=m){if(typ==0) {if(a[xx][yy]=='X') {return false;}}if(typ==1) {if(b[xx][yy]=='X') {return false;}}}}return true;
}int calc()
{kx.clear(); ky.clear();for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){if(a[i][j]=='X') {kx.push_back(i); ky.push_back(j);}}}for(int i=0; i<(1<<kx.size()); i++){for(int j=0; j<kx.size(); j++){if(!(i&(1<<j))) {b[kx[j]][ky[j]] = 'X';}}num[i] = 0;for(int j=1; j<=n; j++){for(int k=1; k<=m; k++){bool ok = check(j,k,1);if((ok && a[j][k]!='X')) num[i]++;}}num[i] += bitcnt[i];for(int j=0; j<kx.size(); j++) {b[kx[j]][ky[j]] = '.';}}dp[0][0] = 1;for(int i=0; i<n*m; i++){for(int j=0; j<(1<<kx.size()); j++){if(dp[i][j]){dp[i+1][j] = (dp[i+1][j]+dp[i][j]*(num[j]-i))%P;for(int k=0; k<kx.size(); k++){if(!(j&(1<<k))){dp[i+1][j|(1<<k)] = (dp[i+1][j|(1<<k)]+dp[i][j])%P;}}}}}int ret = dp[n*m][(1<<kx.size())-1];for(int i=0; i<=n*m; i++) for(int j=0; j<(1<<kx.size()); j++) dp[i][j] = 0;return ret;
}void dfs(int x,int y,int dep)
{if(x==n+1){int tmp = calc();if((dep-cnt)&1) {ans = ans-tmp<0 ? ans-tmp+P : ans-tmp;}else {ans = ans+tmp>=P ? ans+tmp-P : ans+tmp;}return;}int xx = x,yy = y+1; if(yy>m) {yy = 1; xx++;}if(a[x][y]=='X'){bool f = check(x,y,0);if(f) {dfs(xx,yy,dep+1);}}else{a[x][y] = 'X';bool f = check(x,y,0);if(f) {dfs(xx,yy,dep+1);}a[x][y] = '.';dfs(xx,yy,dep);}
}int main()
{for(int i=1; i<(1<<8); i++) bitcnt[i] = bitcnt[i>>1]+(i&1);scanf("%d%d",&n,&m);for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) b[i][j] = '.';for(int i=1; i<=n; i++) scanf("%s",a[i]+1);for(int i=1; i<=n; i++){for(int j=1; j<=m; j++) {if(a[i][j]=='X') cnt++;}}if(cnt>8) {printf("0"); return 0;}dfs(1,1,0);printf("%d\n",ans);return 0;
}