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SRM 602 D2L3:BlackBoxDiv2,dp

2014-02-21 16:44 239 查看
题目:http://community.topcoder.com/stat?c=problem_statement&pm=12929&rd=15820

参考:http://apps.topcoder.com/wiki/display/tc/SRM+602

需要仔细分析,要注意base case的判断,还要注意不要漏了repaint的情况。

代码:

#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <bitset>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>

using namespace std;

#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)
typedef pair<int, int> pii;
typedef long long llong;
typedef pair<llong, llong> pll;
#define mkp make_pair

/*************** Program Begin **********************/
const int MOD = 1e9 + 7;

long long dp[55][55];
long long C[55][55];

class BlackBoxDiv2 {
public:
int w, h;
void calc()
{
C[0][0] = 1;
for (int i = 1; i <= 50; i++) {
C[i][0] = C[i][i] = 1;
for (int j = 1; j < i; j++) {
C[i][j] = C[i-1][j] + C[i-1][j-1];
C[i][j] %= MOD;
}
}
}
long long rec(int x, int y)
{
if (0 == x) {		// base case
if (0 == y) {
return 1;
} else {
return 0;
}
}

long long & res = dp[x][y];
if (res != -1) {
return res;
}

res = 0;
for (int s = 0; s <= y; s++) {
for (int r = 0; r <= h - y; r++) {
if (s + r < 1) {
continue;
}
long long t = C[y][s] * C[h - y][r] % MOD;
res += ( t * rec(x - 1, y - s) ) % MOD;
res %= MOD;
}
}
return res;
}
int count(string front, string side) {
this->w = 0; this->h = 0;
for (int i = 0; i < front.size(); i++) {
w += ( front[i] == 'B' ? 1 : 0 );
}
for (int i = 0; i < side.size(); i++) {
h += ( side[i] == 'B' ? 1 : 0 );
}
calc();
memset(dp, -1, sizeof(dp));
return rec(w, h);
}

};

/************** Program End ************************/
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