[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[210][210],a[210];//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=0;i<n;i++) cin>>a[i];//输入
    memset(sum,0,sizeof(sum));//清0
    ans=-5000000;//最值
    for(i=0;i<n;i++){
        sum[i][i]=a[i];
        if(sum[i][i]>ans) ans=sum[i][i];
        for(j=i+1;j<n;j++){
            sum[i][j]=sum[i][j-1]+a[j];
            if(sum[i][j]>ans) ans=sum[i][j];
        }
    }
    priority_queue<int> temp;
    multiset<int> add;
    for(i=0;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]<0) temp.push(a[o]);
                if(temp.size()>k) temp.pop();
            }
            if((j-i+1)==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-1;o>=0;o--){
                if(a[o]>0) add.insert(a[o]);
                if(add.size()>k) add.erase(add.begin());
            }
            for(o=j+1;o<n;o++){
                if(a[o]>0) 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<<"
";
    return 0;
}