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CF243 DIV2 C Sereja and Swaps(暴力)

2024-07-01 23:36:17   225人阅读

题意:给你一个数组,问你交换最多K个数以后,最大子串和为多少;

解题思路:枚举这个数组最大字串和的起点和终点,然后优先队列交换这段里面的小数去换外面的大数,即可求出答案!

解题代码:

 1 // File Name: c.cpp
 2 // Author: darkdream
 3 // Created Time: 2014年04月28日 星期一 19时04分36秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 #include<queue>
25 using namespace std;
26 
27 
28 struct  cmp1{
29   bool operator()(int x,int y)
30   {
31       return x > y ; 
32   }
33 };
34 struct  cmp2{
35   bool operator()(int x,int y)
36   {
37       return x < y ; 
38   }
39 };
40 int a[300];
41 int main(){
42   
43     
44   int n , k ; 
45   scanf("%d %d",&n,&k);
46   for(int i =1 ;i <= n;i ++)
47   {
48      scanf("%d",&a[i]);
49   }
50   int ans = a[1];
51   priority_queue<int,vector<int>,cmp1> IN;
52   priority_queue<int,vector<int>,cmp2> OUT;
53   int sum[300];
54   memset(sum,0,sizeof(sum));
55   sum[1] = a[1];
56   for(int i = 2;i <= n;i ++ )
57   {
58      sum[i] = sum[i-1] + a[i]; 
59   }
60   for(int begin = 1; begin <= n; begin ++  )
61      for(int end = begin + 1 ; end <= n;end ++ )
62      {
63          int tsum = sum[end] - sum[begin-1];
64          ans = max(tsum,ans); 
65          for(int i = begin;i <= end;i++)
66          {
67             IN.push(a[i]);
68          }
69          for(int i = 1;i < begin ;i ++)
70             OUT.push(a[i]);
71          for(int i = end+1 ; i <= n  ;i ++)
72             OUT.push(a[i]);
73          
74          for(int i = 1;i <= k;i ++)
75          {
76             if(IN.empty() || OUT.empty())
77                 break;
78             int t1 = IN.top();
79             int t2 = OUT.top();
80             IN.pop();
81             OUT.pop();
82             tsum += t2 -t1; 
83             ans = max(tsum,ans);
84          }
85          while(!IN.empty())  IN.pop();
86          while(!OUT.empty())  OUT.pop();
87      }
88      printf("%d\n",ans);
89 return 0;
90 }
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