山东省第七届ACM F Feed the monkey (记忆搜索)

/*题意:3种水果,各有a,b,c个,一天喂一种水果,每种水果不能喂连续的n1,n2,n3天,问有多少种喂的方式思路:简单的记忆搜索dp[i][j][k][s][f]表示第一种水果吃掉i个第二种水果吃掉j个，第三种水果剩余k个，以s种水果结尾连续数量 f 在规定范围内的组合种类*/代码:#include <bits/stdc++.h>using namespace std;#define ll long longconst long long maxn = 52;#define mod 1000000007ll dp[maxn][maxn][maxn][4][maxn];int N[4], D[4], sum;int dfs(int a, int b, int c, int t, int d){    if (a > N[1] || b > N[2] || c > N[3] || a + b + c > sum) return 0;    for (int i = 1; i <= 3; ++i) {        if (t == i && d > D[i]) {            return 0;        }    }    if(dp[a][b][c][t][d] != -1) return dp[a][b][c][t][d];    if (a + b + c == sum) return 1;    int ans = 0;    if(t == 1) {        ans = (ans + dfs(a+1,b,c,1,d + 1)) % mod;        ans = (ans + dfs(a,b+1,c,2,1)) % mod;        ans = (ans + dfs(a,b,c+1,3,1)) % mod;        return dp[a][b][c][t][d]=ans;    }    if(t == 2) {        ans = (ans + dfs(a+1,b,c,1,1)) % mod;        ans = (ans + dfs(a,b+1,c,2,d + 1)) % mod;        ans = (ans + dfs(a,b,c+1,3,1)) % mod;        return dp[a][b][c][t][d]=ans;    }    if(t == 3) {        ans = (ans + dfs(a+1,b,c,1,1)) % mod;        ans = (ans + dfs(a,b+1,c,2,1)) % mod;        ans = (ans + dfs(a,b,c+1,3,d + 1)) % mod;        return dp[a][b][c][t][d]=ans;    }}int main(){    int Test;    scanf("%d", &Test);    while (Test--) {        memset(dp, -1, sizeof(dp));        scanf("%d%d%d%d%d%d", &N[1], &N[2], &N[3], &D[1], &D[2], &D[3]);        int ans = 0;        sum = N[1] + N[2] + N[3];        ans = (ans + dfs(1, 0, 0, 1, 1)) % mod;       // cout << ans << endl;        ans = (ans + dfs(0, 1, 0, 2, 1)) % mod;       // cout << ans << endl;        ans = (ans + dfs(0, 0, 1, 3, 1)) % mod;        cout << ans << endl;    }    return 0;}