• Codeforces 425A Sereja and Swaps(暴力枚举)


    题目链接:A. Sereja and Swaps

    题意:给定一个序列,能够交换k次,问交换完后的子序列最大值的最大值是多少

    思路:暴力枚举每一个区间,然后每一个区间[l,r]之内的值先存在优先队列内,然后找区间外假设有更大的值就替换掉。

    求出每一个区间的最大值,最后记录下全部区间的最大值

    代码:

    By lab104_yifan, contest: Codeforces Round #243 (Div. 2), problem: (C) Sereja and Swaps, Accepted, #
     #include <stdio.h>
    #include <string.h>
    #include <queue>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define max(a, b) ((a)>(b)?(a):(b))
    int n, k, a[205], i, j, vis[205];
    
    struct cmp {
        bool operator() (int &a, int &b) {
            return a > b;
        }
    };
    
    int cal(int l, int r) {
        priority_queue<int, vector<int>, cmp> Q;
        int sum = 0, i;
        for (i = l; i <= r; i++) {
            sum += a[i];
            Q.push(a[i]);
        }
        int kk = k;
        memset(vis, 0, sizeof(vis));
        while (kk--) {
            int maxx = -INF, max_v;
            for (i = 0; i < n; i++) {
                if ((i >= l && i <= r) || vis[i]) continue;
                if (maxx < a[i]) {
                    maxx = a[i];
                    max_v = i;
                }
            }
            if (Q.top() < maxx) {
                sum = sum - Q.top() + maxx;
                Q.pop();
                Q.push(maxx);
                vis[max_v] = 1;
            }
        }
        return sum;
    }
    
    int main() {
        int ans = -INF;
        scanf("%d%d", &n, &k);
        for (i = 0; i < n; i++)
            scanf("%d", &a[i]);
        for (i = 0; i < n; i++)
            for (j = i; j < n; j++) {
                int t = cal(i, j);
                ans = max(ans, t);
            }
        printf("%d
    ", ans);
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/slgkaifa/p/6984231.html
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