HDU 4804 Campus Design(经典轮廓线问题扩展)



题意：给定一个图，0是不能放的，然后现在有1X1和1X2方块，最后铺满该图，使得1X1使用次数在C到D之间，1X2次数随便，问有几种放法。

思路：经典问题多米诺的拓展，状态多开一维表示用了几个1*1砖块即可，注意位运算的优先级。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#define eps 1e-6
#define LL long long
#define pii (pair<int, int>)
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
int n, m, cur, C, D;
const int maxn = 11;
LL d[2][1<<maxn][25];
char G[150][maxn];

int main() {
//	freopen("input.txt", "r", stdin);
	while(scanf("%d%d%d%d", &n, &m, &C, &D) == 4) {
		for(int i = 0; i < n; i++) {
			getchar();
			for(int j = 0; j < m; j++) scanf("%c", &G[i][j]);
		}
		cur = 0;
		memset(d[cur], 0, sizeof(d[cur]));
		d[cur][(1<<m)-1][0] = 1;
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++) {
				cur ^= 1;
				memset(d[cur], 0, sizeof(d[cur]));
				for(int k = 0; k < (1<<m); k++) {
					if(G[i][j] == '1') {
						if((k&(1<<(m-1))) || (i&&G[i-1][j]=='0')) for(int l = 0; l <= D; l++)
								d[cur][(k<<1)&(~(1<<m))][l] += d[1-cur][k][l], d[cur][(k<<1)&(~(1<<m))][l] %= mod; //不放
						if(i && !(k&(1<<(m-1))) && G[i-1][j]!='0') for(int l = 0; l <= D; l++)
								d[cur][(k<<1)+1][l] += d[1-cur][k][l], d[cur][(k<<1)+1][l] %= mod; //竖放1*2
						if(j && G[i][j-1]!='0' && !(k&1) && ((k&(1<<(m-1)))||i&&G[i-1][j]=='0')) for(int l = 0; l <= D; l++)
								d[cur][(k<<1)&(~(1<<m))|3][l] += d[1-cur][k][l], d[cur][(k<<1)&(~(1<<m))|3][l] %= mod; //右放1*2
						if((k&(1<<(m-1))) || i&&G[i-1][j]=='0') for(int l = 0; l < D; l++)
								d[cur][((k<<1)&(~(1<<m)))+1][l+1] += d[1-cur][k][l], d[cur][((k<<1)&(~(1<<m)))+1][l+1] %= mod; //放1*1
					}
					else {
						if((k&(1<<(m-1))) || i&&G[i-1][j]=='0') for(int l = 0; l <= D; l++)
							d[cur][(k<<1)&(~(1<<m))][l] += d[1-cur][k][l], d[cur][(k<<1)&(~(1<<m))][l] %= mod;
					}
				}
			}
		LL ans = 0;
		int s = (1<<m)-1;
		for(int i = 0; i < m; i++) if(G[n-1][i] == '0') s ^= (1<<(m-i-1));
		//cout << s << endl;
		for(int i = C; i <= D; i++) ans += d[cur][s][i], ans %= mod;
		//cout << d[1-cur][s][0] << endl;
		cout << ans << endl;
		//for(int i = C; i <= D; i++) cout << d[1-cur][2][0] << endl;
		//cout << G[0][0] << G[0][1] << endl;
	}
	return 0;
}

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