[ACM_暴力] 最多交换k个数的顺序，求a[i]的最大连续和

 

 

 1 /*

 2 http://codeforces.com/contest/426/problem/C

 3 最多交换k个数的顺序，求a[i]的最大连续和

 4 爆解

 5 思路：Lets backtrack interval that should contain maximal sum. 

 6 To improve it we can swap not more then K minimal elements

 7  in fixed interval to not more K maximal elements not 

 8  contained in our interval. As n is quite little we can 

 9  do it in any way. Author solution works O(n3?*?log(n)).

10 */

11 #include <iostream>

12 #include <vector>

13 #include <string.h>

14 #include <algorithm>

15 #include <queue>

16 #include <set>

17 

18 using namespace std;

19 int sum[210][210],a[210];//sum[i][j]保存从i到j的和

20 int main(){

21     int n,k,i,j,o,ans;

22     cin>>n>>k;

23     for(i=0;i<n;i++) cin>>a[i];//输入

24     memset(sum,0,sizeof(sum));//清0

25     ans=-5000000;//最值

26     for(i=0;i<n;i++){

27         sum[i][i]=a[i];

28         if(sum[i][i]>ans) ans=sum[i][i];

29         for(j=i+1;j<n;j++){

30             sum[i][j]=sum[i][j-1]+a[j];

31             if(sum[i][j]>ans) ans=sum[i][j];

32         }

33     }

34     priority_queue<int> temp;

35     multiset<int> add;

36     for(i=0;i<n;i++){

37         for(j=i;j<n;j++){

38             for(o=i;o<=j;o++){//查找从i到j为负的最小的k个

39                 if(a[o]<0) temp.push(a[o]);

40                 if(temp.size()>k) temp.pop();

41             }

42             if((j-i+1)==temp.size()){//全为负

43                 sum[i][j]=temp.top();

44                 while(!temp.empty()) temp.pop();//清空

45             }

46             else{

47                 while(!temp.empty()){//清空

48                     sum[i][j]-=temp.top();//将负的移出

49                     temp.pop();

50                 }

51             }

52 

53             add.clear();//清空add

54             //对于非[i,j]部分取k个最大正数

55             for(o=i-1;o>=0;o--){

56                 if(a[o]>0) add.insert(a[o]);

57                 if(add.size()>k) add.erase(add.begin());

58             }

59             for(o=j+1;o<n;o++){

60                 if(a[o]>0) add.insert(a[o]);

61                 if(add.size()>k) add.erase(add.begin());

62             }

63             while(!add.empty()){

64                 sum[i][j]+=*add.begin();

65                 add.erase(add.begin());

66             }

67             if(sum[i][j]>ans) ans=sum[i][j];//更新ans

68         }

69     }

70     cout<<ans<<"\n";

71     return 0;

72 }