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\n",++cse);
continue;
}
ans=0;
dfs(0,0);
if(ans<0) ans+=772002;
printf("Case #%d: %lld\n",++cse,ans);
}
return 0;
}
