Codeforces 425A Sereja and Swaps（暴力枚舉）

題目連結：A. Sereja and Swaps

題意：給定一個序列，可以交換k次，問交換完後的子序列最大值的最大值是多少

思路：暴力枚舉每一個區間，然後每個區間[l,r]之內的值先存在優先隊列內，然後找區間外如果有更大的值就替換掉。求出每個區間的最大值，最後記錄下所有區間的最大值

代碼：

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\n", ans);    return 0;}