题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4804
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:插头DP的变形,只要多考虑1X1的情况即可,然后DP多开一维表示使用1X1的个数
代码:
#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; }
HDU 4804 Campus Design(插头DP),布布扣,bubuko.com
原文地址:http://blog.csdn.net/accelerator_/article/details/26097779