hdu-5583 Kingdom of Black and White(数学,贪心,暴力)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5583
Kingdom of Black and White
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Now N frogs are standing in a line, some of them are black, the others are white. The total strength of those frogs are calculated by dividing the line into minimum parts, each part should still be continuous, and can only contain one kind of frog. Then the strength is the sum of the squared length for each part.
However, an old, evil witch comes, and tells the frogs that she will change the color of at most one frog and thus the strength of those frogs might change.
The frogs wonder the maximum possible strength after the witch finishes her job.
Every test case only contains a string with length N, including only 0 (representing
a black frog) and 1 (representing a white frog).
⋅ 1≤T≤50.
⋅ for 60% data, 1≤N≤1000.
⋅ for 100% data, 1≤N≤105.
⋅ the string only contains 0 and 1.
000011
0101
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=1e5+;
long long dp[N];
char str[N];
int main()
{
int t;
scanf("%d",&t);
int cnt=;
while(t--)
{
int num=;
long long ans=,sum=;
scanf("%s",str);
int len=strlen(str);
memset(dp,,sizeof(dp));
dp[]=;
for(int i=;i<len;i++)
{
if(str[i]==str[i-])
{
dp[num]++;
}
else
{
sum+=dp[num]*dp[num];
num++;
dp[num]++;
}
}
sum+=dp[num]*dp[num];
if(num==)
{
ans=dp[]*dp[];
}
for(int i=;i<=num;i++)
{
if(dp[i]==)
{
ans=max(ans,sum+*(dp[i-]*dp[i+]+dp[i-]+dp[i+]));
}
else
{
if(dp[i-]>=dp[i])
{
ans=max(ans,sum+*(dp[i-]-dp[i]+));
}
else ans=max(ans,sum+*(dp[i]-dp[i-]+));
}
}
cout<<"Case #"<<cnt++<<": "<<ans<<"\n";
}
return ;
}
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