2015
09-18

Campus Design

Nanjing University of Science and Technology is celebrating its 60th anniversary. In order to make room for student activities, to make the university a more pleasant place for learning, and to beautify the campus, the college administrator decided to start construction on an open space.
The designers measured the open space and come to a conclusion that the open space is a rectangle with a length of n meters and a width of m meters. Then they split the open space into n x m squares. To make it more beautiful, the designer decides to cover the open space with 1 x 1 bricks and 1 x 2 bricks, according to the following rules:

1. All the bricks can be placed horizontally or vertically
2. The vertexes of the bricks should be placed on integer lattice points
3. The number of 1 x 1 bricks shouldn’t be less than C or more than D. The number of 1 x 2 bricks is unlimited.
4. Some squares have a flowerbed on it, so it should not be covered by any brick. (We use 0 to represent a square with flowerbet and 1 to represent other squares)

Now the designers want to know how many ways are there to cover the open space, meeting the above requirements.

There are several test cases, please process till EOF.
Each test case starts with a line containing four integers N(1 <= N <= 100), M(1 <= M <= 10), C, D(1 <= C <= D <= 20). Then following N lines, each being a string with the length of M. The string consists of ‘0’ and ‘1’ only, where ‘0’ means the square should not be covered by any brick, and ‘1’ otherwise.

There are several test cases, please process till EOF.
Each test case starts with a line containing four integers N(1 <= N <= 100), M(1 <= M <= 10), C, D(1 <= C <= D <= 20). Then following N lines, each being a string with the length of M. The string consists of ‘0’ and ‘1’ only, where ‘0’ means the square should not be covered by any brick, and ‘1’ otherwise.

1 1 0 0
1
1 1 1 2
0
1 1 1 2
1
1 2 1 2
11
1 2 0 2
01
1 2 0 2
11
2 2 0 0
10
10
2 2 0 0
01
10
2 2 0 0
11
11
4 5 3 5
11111
11011
10101
11111

0
0
1
1
1
2
1
0
2
954

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

const int MOD = 1000000007;
int n, m, c, d, pre = 0, now = 1;
long long dp[2][25][1222];
char g[105][15];

int main() {
while (~scanf("%d%d%d%d", &n, &m, &c, &d)) {
int maxs = (1<<m);
memset(dp[now], 0, sizeof(dp[now]));
dp[now][0][maxs - 1] = 1;
for (int i = 0; i < n; i++)
scanf("%s", g[i]);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
swap(pre, now);
memset(dp[now], 0, sizeof(dp[now]));
int tmp = g[i][j] - '0';
if (tmp) {
for (int k = 0; k <= d; k++) {
for (int s = 0; s < maxs; s++) {
if (k && (s&1<<j))
dp[now][k][s] = (dp[now][k][s] + dp[pre][k - 1][s]) % MOD;//放1X1
if (j && !(s&1<<(j-1)) && (s&1<<j))
dp[now][k][s|1<<(j-1)] = (dp[now][k][s|1<<(j-1)] + dp[pre][k][s]) % MOD;//横放1X2
dp[now][k][s^1<<j] = (dp[now][k][s^1<<j] + dp[pre][k][s]) % MOD;//竖放1X2
}
}
}
else {
for (int k = 0; k <= d; k++) {
for (int s = 0; s < maxs; s++) {
if ((s&1<<j))
dp[now][k][s] = (dp[now][k][s] + dp[pre][k][s]) % MOD;//不能放的情况，和放1X1类似
}
}
}
}
}
long long ans = 0;
for (int i = c; i <= d; i++)
ans = (ans + dp[now][i][maxs - 1]) % MOD;
printf("%lld\n", ans);
}
return 0;
}