2016中国大学生程序设计竞赛 - 网络选拔赛 1007 Mountain hdu5838

Problem Description

Zhu found a map which is a N∗M rectangular grid.Each cell has a height and there are no two cells which have the same height. But this map is too old to get the clear information,so Zhu only knows cells which are valleys.

A cell is called valley only if its height is less than the heights of all its adjacent cells.If two cells share a side or a corner they are called adjacent.And one cell will have eight adjacent cells at most.

Now give you N strings,and each string will contain M characters.Each character will be '.' or uppercase 'X'.The j-th character of the i-th string is 'X' if the j-th cell of the i-th row in the mountain map is a valley, and '.' otherwise.Zhu wants you to calculate the number of distinct mountain maps that match these strings.

To make this problem easier,Zhu defines that the heights are integers between 1 and N∗M.Please output the result modulo 772002.

 

Input

The input consists of multiple test cases. 

Each test case begins with a line containing two non-negative integers N and M. Then N lines follow, each contains a string which contains M characters. (1≤N≤5,1≤M≤5).

 

Output

For each test case, output a single line "Case #x: y", where x is the case number, starting from 1. And y is the answer after module 772002.

 

Sample Input

2 4.X.....X4 2X.....X.1 2XX

 

Sample Output

Case #1: 2100Case #2: 2520Case #3: 0

CQOI2012原题。听说UESTC的验题人一个都没看出是原题

---------------------------------------------------------------------------------------------------

首先因为n<=5,m<=5，所以X的数量不可能超过9个

然后我们就可以状压了。f[i][S]表示填了前I个数，X的位置状态为S的方案数

S表示填或者没填

转移方程就是

f[i][j]=f[i-1][j]*(cnt[j]-i+1)+sigma(f[i][j]+f[i-1][j^(1<<(k-1))])

其中cnt[j]表示在状态j的时候，可以填的格子数量。

因为我们是从小往大填数的，所以有了上述转移式

题目又要求,的地方一定不能是区域最小值

所以我们要枚举可能成为最小值的位置。然后容斥原理处理一下就可以了

#include<map>#include<cmath>#include<queue>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct point{int x,y;}p[10],pp[10];long long mod=772002;long long f[26][2001],cnt[2001];int mx[6][6],a[6][6];bool v[7][7];int n,m,tot,px;int dx[9]={-1,-1,-1,0,0,1,1,1,0};int dy[9]={-1,0,1,-1,1,-1,0,1,0};inline void prepare(){memset(cnt,0,sizeof(cnt));int i,j,k;for(k=0;k<=(1<<tot)-1;k++){//if(cnt[k]!=0&&k<=(1<<px)-1&&k!=0)//continue;memset(mx,0,sizeof(mx));for(i=1;i<=tot;i++){if(((1<<(i-1))&k)==0){for(j=0;j<=8;j++){int x=p[i].x+dx[j],y=p[i].y+dy[j];if(x>=1&&x<=n&&y>=1&&y<=m)mx[x][y]=1;}}}int sum=0;for(i=1;i<=n;i++)for(j=1;j<=m;j++)if(mx[i][j]==0)sum++;cnt[k]=sum;}}inline long long dfs(int xx,int yy,int s){if(xx==n+1){prepare();memset(f,0,sizeof(f));f[0][0]=1;int i,j,k;for(i=1;i<=m*n;i++){for(j=0;j<=(1<<tot)-1;j++){f[i][j]=f[i-1][j]*(cnt[j]-i+1)%mod;for(k=1;k<=tot;k++){if((j&(1<<(k-1)))!=0)f[i][j]=(f[i][j]+f[i-1][j^(1<<(k-1))])%mod;}}}if(s%2==1)return -f[n*m][(1<<tot)-1];elsereturn f[n*m][(1<<tot)-1];}long long ans=0;if(v[xx][yy]){int tx=xx,ty=yy;yy++;if(yy>m){xx++;yy=1;}ans+=dfs(xx,yy,s);}else{bool flag=true;int i;for(i=0;i<=8;i++){int x=xx+dx[i],y=yy+dy[i];if(v[x][y]){flag=false;break;}}int tx=xx,ty=yy;yy++;if(yy>m){xx++;yy=1;}if(flag){v[tx][ty]=true;tot++;p[tot].x=tx;p[tot].y=ty;ans=(ans+mod+dfs(xx,yy,s+1))%mod;v[tx][ty]=false;tot--;}ans=(ans+mod+dfs(xx,yy,s))%mod;}return ans;}int main(){int kx=0;while(scanf("%d%d",&n,&m)!=EOF){memset(cnt,0,sizeof(cnt));memset(v,0,sizeof(v));tot=0;px=0;kx++;int i,j,k;string x;for(i=1;i<=n;i++){cin>>x;for(j=1;j<=m;j++){if(x[j-1]=='.')a[i][j]=0;else{tot++;a[i][j]=tot;v[i][j]=true;p[tot].x=i;p[tot].y=j;}}}px=tot;bool flag=true;for(i=1;i<=tot;i++){for(j=0;j<=7;j++){int x=p[i].x+dx[j],y=p[i].y+dy[j];if(v[x][y]){flag=false;break;}}if(!flag)break;}if(flag)printf("Case #%d: %I64d\n",kx,dfs(1,1,0));elseprintf("Case #%d: 0\n",kx);}return 0;}