[ACM_暴力] 最多交换k个数的顺序，求a[i]的最大连续和

 /*
 http://codeforces.com/contest/426/problem/C
 最多交换k个数的顺序，求a[i]的最大连续和
 爆解
 思路：Lets backtrack interval that should contain maximal sum.
 To improve it we can swap not more then K minimal elements
  in fixed interval to not more K maximal elements not
  contained in our interval. As n is quite little we can
  do it in any way. Author solution works O(n3?*?log(n)).
 */
 #include <iostream>
 #include <vector>
 #include <string.h>
 #include <algorithm>
 #include <queue>
 #include <set>

 using namespace std;
 int sum[][],a[];//sum[i][j]保存从i到j的和
 int main(){
     int n,k,i,j,o,ans;
     cin>>n>>k;
     for(i=;i<n;i++) cin>>a[i];//输入
     memset(sum,,sizeof(sum));//清0
     ans=-;//最值
     for(i=;i<n;i++){
         sum[i][i]=a[i];
         if(sum[i][i]>ans) ans=sum[i][i];
         for(j=i+;j<n;j++){
             sum[i][j]=sum[i][j-]+a[j];
             if(sum[i][j]>ans) ans=sum[i][j];
         }
     }
     priority_queue<int> temp;
     multiset<int> add;
     for(i=;i<n;i++){
         for(j=i;j<n;j++){
             for(o=i;o<=j;o++){//查找从i到j为负的最小的k个
                 if(a[o]<) temp.push(a[o]);
                 if(temp.size()>k) temp.pop();
             }
             if((j-i+)==temp.size()){//全为负
                 sum[i][j]=temp.top();
                 while(!temp.empty()) temp.pop();//清空
             }
             else{
                 while(!temp.empty()){//清空
                     sum[i][j]-=temp.top();//将负的移出
                     temp.pop();
                 }
             }

             add.clear();//清空add
             //对于非[i,j]部分取k个最大正数
             for(o=i-;o>=;o--){
                 if(a[o]>) add.insert(a[o]);
                 if(add.size()>k) add.erase(add.begin());
             }
             for(o=j+;o<n;o++){
                 if(a[o]>) add.insert(a[o]);
                 if(add.size()>k) add.erase(add.begin());
             }
             while(!add.empty()){
                 sum[i][j]+=*add.begin();
                 add.erase(add.begin());
             }
             if(sum[i][j]>ans) ans=sum[i][j];//更新ans
         }
     }
     cout<<ans<<"\n";
     return ;
 }