HDU5838 Mountain（状压DP + 容斥原理）

题目

Source

http://acm.hdu.edu.cn/showproblem.php?pid=5838

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

Sample Output

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

分析

题目大概说给一张大小n*m由'X'或'.'标记的图，要在其各个位置填入1到n*m这几个数，满足'X'填入的数是极小的（小于周围八个数）且'.'填入的数不是极小的，求有几种填法。

好难的感觉。。
考虑把数从小到大依次填入：

• dp[i][S]表示当前已经填入数字1...i，且已经被填数的'X'集合为S的方案数（因为合法的'X'最多有9个所以用9位二进制压缩表示S即可）

通过将i+1填入转移到dp[i+1]的状态：

• 可以选择把i+1填入下一个还没被填的'X'位置x
• 也可以填入'.'的位置，这个点'.'的位置不能是未填的'X'周围，因为是从小到大填数的

这样子得到的方案数得到的不是答案要的方案数，因为它还包含了'.'填入的数是极小的，这个用容斥原理搞一下，具体就是用一个dfs搜出所有合法的'.'变为'X'的情况，奇数个减，偶数个加。。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int n,m;

int dx[]={0,0,1,-1,1,-1,1,-1};
int dy[]={1,-1,0,0,1,-1,-1,1};
char map[5][5];

long long d[26][1<<9];
int pos[1<<9],siz[1<<9];

long long dp(){
    int cnt=0,x[9],y[9];
    for(int i=0; i<n; ++i){
        for(int j=0; j<m; ++j){
            if(map[i][j]=='X'){
                x[cnt]=i; y[cnt]=j;
                ++cnt;
            }
        }
    }
    memset(siz,0,sizeof(siz));
    for(int s=0; s<(1<<cnt); ++s){
        for(int i=0; i<cnt; ++i){
            if(s>>i&1){
                map[x[i]][y[i]]='Y';
                ++siz[s];
            }
        }
        int res=0;
        for(int i=0; i<n; ++i){
            for(int j=0; j<m; ++j){
                if(map[i][j]!='.') continue;
                bool flag=1;
                for(int k=0; k<8; ++k){
                    int nx=i+dx[k],ny=j+dy[k];
                    if(nx<0 || nx>=n || ny<0 ||ny>=m) continue;
                    if(map[nx][ny]=='X'){
                        flag=0;
                        break;
                    }
                }
                if(flag) ++res;
            }
        }
        pos[s]=res;
        for(int i=0; i<cnt; ++i){
            if(s>>i&1){
                map[x[i]][y[i]]='X';
            }
        }
    }

    memset(d,0,sizeof(d));
    d[0][0]=1;
    for(int i=0; i<n*m; ++i){
        for(int j=0; j<(1<<cnt); ++j){
            if(d[i][j]==0) continue;
            for(int k=0; k<cnt; ++k){
                if(j>>k&1) continue;
                d[i+1][j^(1<<k)]+=d[i][j];
                d[i+1][j^(1<<k)]%=772002;
            }
            d[i+1][j]+=d[i][j]*(pos[j]-i+siz[j]);
            d[i+1][j]%=772002;
        }
    }
    return d[n*m][(1<<cnt)-1];
}

long long ans;
void dfs(int z,int k){
    if(z==n*m){
        if(k&1) ans-=dp();
        else ans+=dp();
        ans%=772002;
        return;
    }
    int x=z/m,y=z%m;
    bool flag=(map[x][y]!='X');
    for(int i=0; i<8; ++i){
        int nx=x+dx[i],ny=y+dy[i];
        if(nx<0 || nx>=n || ny<0 || ny>=m) continue;
        if(map[nx][ny]=='X'){
            flag=0;
            break;
        }
    }
    if(flag){
        map[x][y]='X';
        dfs(z+1,k+1);
        map[x][y]='.';
    }
    dfs(z+1,k);
}

bool check(){
    for(int i=0; i<n; ++i){
        for(int j=0; j<m; ++j){
            if(map[i][j]=='.') continue;
            for(int k=0; k<8; ++k){
                int nx=i+dx[k],ny=j+dy[k];
                if(nx<0 || nx>=n || ny<0 || ny>=m) continue;
                if(map[nx][ny]=='X') return 0;
            }
        }
    }
    return 1;
}

int main(){
    int cse=0;
    while(~scanf("%d%d",&n,&m)){
        for(int i=0; i<n; ++i){
            for(int j=0; j<m; ++j){
                scanf(" %c",&map[i][j]);
            }
        }
        if(!check()){
            printf("Case #%d: 0
",++cse);
            continue;
        }
        ans=0;
        dfs(0,0);
        if(ans<0) ans+=772002;
        printf("Case #%d: %lld
",++cse,ans);
    }
    return 0;
}