hdu 5890 01背包 bitset

注意不能每个T都mem 不然会T

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double EPS = 1.0e-8;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 205;
const int  maxm = 300;
//next_permutation
int t;
int n;
int num[105];
bitset<100> f[15];
int dp[55][55][55];
int ques[5];
bool ok(int a, int b, int c, int x)
{
        if (x == a || x == b || x == c)
        {
                return true;
        }
        return false;
}
void getans()
{
        for (int i = 1; i <= n; i++)
                for (int j = i; j <= n; j++)
                        for (int k = j; k <= n; k++)
                        {
                                for (int l = 1; l <= 10; l++)
                                {
                                        f[l].reset();
                                }
                                f[0][0] = 1;
                                for (int l = 1; l <= n; l++)
                                {
                                        for (int p = 10; p >= 1; p--)
                                        {
                                                if (ok(i, j, k, l) || num[l] > 87)
                                                {
                                                        continue;
                                                }
                                                f[p] |= f[p - 1] << num[l];
                                        }
                                }
                                if (f[10][87])
                                {
                                        dp[i][j][k] = 1;
                                }
                                else
                                {
                                        dp[i][j][k] = 0;
                                }
                        }
}
int main()
{
        cin >> t;
        while (t--)
        {
                //mem(dp, 0);
                scanf("%d", &n);
                for (int i = 1; i <= n; i++)
                {
                        scanf("%d", &num[i]);
                }
                getans();
                int m;
                int a, b, c;
                scanf("%d", &m);
                for (int i = 1; i <= m; i++)
                {
                        scanf("%d%d%d", &ques[1], &ques[2], &ques[3]);
                        sort(ques + 1, ques + 4);
                        if (dp[ques[1]][ques[2]][ques[3]])
                        {
                                printf("Yes
");
                        }
                        else
                        {
                                printf("No
");
                        }
                }
        }
}
原文地址:https://www.cnblogs.com/Aragaki/p/7683194.html