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，可以交换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，然后每个区间[l,r]之内的值先存在优先队列内，然后找区间外如果有更大的值就替换掉。求出每个区间的最大值，最后记录下所有区间的最大值

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\n", ans);
return 0;
}