暴力水题（思路很重要）HDU5583

Kingdom of Black and White

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 243 Accepted Submission(s): 98



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）

#include<stdio.h>
#include<string.h>
long long max(long long a,long long b)
{
if(a<b)
return  b;
else return a;
}
int  main()
{
long long test;
int sb=1;
scanf("%I64d",&test);
while(test--)
{
char a[100005];
long long map[100005];

memset(map,0,sizeof(map));
scanf("%s",a);
long long la=strlen(a);
long long count=0;
long long res=0;
for(long long i=0; i<la;)
{
long long j=i;
while(j<la&&a[i]==a[j])
j++;
map[count]=j-i;
res+=map[count]*map[count];
count++;
i=j;
}
//        prlong longf("count:%I64d\n",count);
//        for(long long  i=0; i<count; i++)
//            prlong longf("%I64d\n",map[i]);
if(count==1)
{
printf("Case #%d: %I64d\n",sb++,res);
continue;
}
long long ans=res;
for(long long i=0; i<count; i++)//当长度为1是会对左右都产生影响，不为1则没有影响
{

long long  t=res;
if(map[i]==1)
{
long long  now=map[i];
t-=map[i]*map[i];
if(i<count-1)
{
now+=map[i+1];
t-=map[i+1]*map[i+1];
}
if(i)
{
now+=map[i-1];
t-=map[i-1]*map[i-1];
}
t+=now*now;
ans=max(ans,t);
}
else
{

if(i<count-1)
{
long long temp=res;
temp-=map[i]*map[i];
temp-=(map[i+1])*(map[i+1]);
temp+=(map[i+1]+1)*(map[i+1]+1);
temp+=(map[i]-1)*(map[i]-1);
ans=max(ans,temp);
}
if(i)
{
long long temp=res;
temp-=map[i]*map[i];
temp-=(map[i-1])*(map[i-1]);
temp+=(map[i-1]+1)*(map[i-1]+1);
temp+=(map[i]-1)*(map[i]-1);
ans=max(ans,temp);
}

}
}
printf("Case #%d: %I64d\n",sb++,ans);

}
}