【动态规划】TopCoder SRM 573 division2 WolfPackDivTwo

比赛的时候没做出来，当时的思路是计算组合数，这个思路应该也没错，但是有一些情况我不会求。比如走m步回到原地的方案数，到现在也没想明白……

参考答案是动态规划的方式，将原问题分解成一个个小的问题，避开了求组合数，同时利用备忘录的方法大大减小了时间复杂度。

我的代码：

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define MOD 1000000007

using namespace std;

const int dir_x[] = {0, 1, 0, -1};
const int dir_y[] = {1, 0, -1, 0};
// Divide original problem into smaller problems
// And then using dynamic programming algorithm
class WolfPackDivTwo {
public:
// dx, dy, m
int f[55][55][55];
int calc(vector <int>, vector <int>, int);

// Get the result of dx, dy, m
int get(int dx, int dy, int m){
// Boundary conditions
if(m == 0){
if(dx==0 && dy==0)
return 1;
else return 0;
//	return (dx==0 && dy==0);
}
if(f[dx][dy][m] == -1){
f[dx][dy][m] = 0;
for(int i=0; i<4; i++){
int tx = abs(dx+dir_x[i]);
int ty = abs(dy+dir_y[i]);
if(tx+ty <= m-1)
f[dx][dy][m] = (f[dx][dy][m]+get(tx, ty, m-1))%MOD;
}
}
return f[dx][dy][m];
}

};

int WolfPackDivTwo::calc(vector <int> x, vector <int> y, int m) {
int length = x.size();
long long res = 0;

// Initialization
for(int i=0; i<55; i++)
for(int j=0; j<55; j++)
for(int k=0; k<55; k++)
f[i][j][k] = -1;

// For every position
for(int i=-51; i<105; i++)
for(int j=-51; j<105; j++){
long long tmp = 1;
for(int k=0; k<length; k++){
if(abs(x[k]-i)+abs(y[k]-j) > m){
tmp = 0;
break;
}
else{
tmp = (tmp*get(abs(x[k]-i), abs(y[k]-j), m))%MOD;
}
}
res = (res+tmp)%MOD;
}
return res;
}

//<%:testing-code%>
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