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  • HDU 5691 Sitting in Line

    状态压缩DP。

    dp[i][j]表示前cnt个位置放了i状态的那些数字,cnt位置放的是j这个数字的最大价值。其中cnt为i二进制中1的个数。

    #include<cstdio>
    #include <iostream>
    #include<cstring>
    #include<cmath>
    #include<vector>
    #include<queue>
    #include<algorithm>
    using namespace std;
    
    long long INF = 99999999999999999;
    long long dp[70000][20];
    int n;
    long long a[20];
    int p[20],f[20], g[20];
    
    int main()
    {
        int T; int Case = 1;
        scanf("%d", &T);
        while (T--)
        {
            scanf("%d", &n);
            memset(f, 0, sizeof f); memset(g, 0, sizeof g);
            for (int i = 1; i <= n; i++)
            {
                scanf("%lld%d", &a[i], &p[i]);
                if (p[i] == -1) continue; 
                p[i]++;
                f[p[i]] = i;
                g[i] = 1;
            }
    
            for (int i = 0; i<(1 << n); i++)
                for (int j = 0; j <= n; j++) dp[i][j] = -INF;
    
            for (int j = 1; j <= n; j++)
            {
                if (g[j] == 1 && p[j] != 1) continue;
                dp[1 << (j - 1)][j] = 0;
            }
    
            for (int i = 1; i<(1 << n); i++)
            {
                int cnt = 0;
                for (int j = 0; j<n; j++) if ((i&(1 << j)) !=0) cnt++;
    
                for (int j = 1; j <= n; j++)
                {
                    if (dp[i][j] == -INF) continue;
                    if (f[cnt + 1] != 0)
                    {
                        dp[i | (1 << (f[cnt + 1] - 1))][f[cnt + 1]] = max(dp[i | (1 << (f[cnt + 1] - 1))][f[cnt + 1]],
                            dp[i][j] + a[f[cnt + 1]] * a[j]);
                        continue;
                    }
                    for (int k = 1; k <= n; k++)
                    {
                        if (((1 << (k - 1))&i) != 0) continue;
                        if (g[k] == 1) continue;
    
                        dp[i | (1 << (k - 1))][k] = max(dp[i | (1 << (k - 1))][k],
                            dp[i][j] + a[k] * a[j]);
                    }
                }
            }
    
            long long ans = -INF;
            for (int j = 1; j <= n; j++)
            {
                ans = max(ans, dp[(1 << n) - 1][j]);
            }
            printf("Case #%d:
    ", Case++);
            printf("%lld
    ", ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zufezzt/p/5519087.html
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