hdu 5583 Kingdom of Black and White【枚举】

Kingdom of Black and White

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1323    Accepted Submission(s): 407


Problem Description

In the Kingdom of Black and White (KBW), there are two kinds of frogs: black frog and white frog.

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.

Input

First line contains an integer T, which indicates the number of test cases.

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.

Output

For every test case, you should output "Case #x: y",where x indicates the case number and counts from 1 and y is the answer.

Sample Input

2

000011

0101

Sample Output

Case #1: 26

Case #2: 10

Source

2015ACM/ICPC亚洲区上海站-重现赛（感谢华东理工）

 

 题目大意：两种颜色的青蛙站一排，连在一起的n个青蛙贡献的值为n*n，比如第一组样例：4*4+2*2=20，现在我们可以任意让一个青蛙变色，问改变一只青蛙的颜色之后的最大值。

思路：

1、将每个区间的青蛙数保存在一个数组b【】中，b【i】表示第i个连在一起的青蛙区间中有多少青蛙。比如第一组样例：b【0】=4，b【1】=2；

2、暴力枚举每个区间：

如果区间青蛙数大于1：

①让区间的左边少一只青蛙，让其和左边的区间连在一起。

②让区间的右边少一只青蛙，让其和右边的区间连在一起。

否则：

连接其左边的区间和右边的区间和这只青蛙。

并维护最大值即可

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long int
char a[1000000];
ll b[1000000];
int main()
{
int t;
int kase=0;
scanf("%d",&t);
while(t--)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
scanf("%s",a);
int n=strlen(a);
int cont=0;
for(int i=0;i<n;i++)
{
for(int j=i;j<n;j++)
{
if(a[i]!=a[j])
{
b[cont++]=j-i;
i=j-1;
break;
}
if(j==n-1)
{
b[cont++]=j-i+1;
i=n;
break;
}
}
}
ll output=0;
for(int i=0;i<cont;i++)
{
output+=b[i]*b[i];
}
ll pre=output;
for(int i=0;i<cont;i++)
{
if(b[i]-1==0)
{
ll y=0,z=0;
if(i-1>=0)y=b[i-1];
if(i+1<cont)z=b[i+1];
ll tmp=pre-1-y*y-z*z+(1+y+z)*(1+y+z);
output=max(output,tmp);
continue;
}
ll tmp=pre-b[i]*b[i]+(b[i]-1)*(b[i]-1);
if(i-1>=0)
output=max(tmp+(b[i-1]+1)*(b[i-1]+1)-(b[i-1]*(b[i-1])),output);
if(i+1<cont)
output=max(tmp+(b[i+1]+1)*(b[i+1]+1)-(b[i+1]*b[i+1]),output);
}
printf("Case #%d: %I64d\n",++kase,output);
}
}