HDU 5583 Kingdom of Black and White

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5583

题意：一个01序列的值为连续相同元素作为一个部分，各部分元素个数的平方之和。现在最多修改一个元素（0变1,1变0），求这个序列的最大值。

思路：因为最多只变一个元素，所以我们先统计出每个部分的元素个数，然后直接枚举取最大值即可，注意当一个部分只有一个元素时，会把三部分接起来。

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <iostream>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <sstream>#include <queue>#include <utility>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)#define Rrep(i,j,k) for (int i=j;i>=k;i--)#define Clean(x,y) memset(x,y,sizeof(x))#define LL long long#define ULL unsigned long long#define inf 0x7fffffff#define mod 100000007const int maxn = 100009;int T;char str[maxn];LL f[maxn];LL num[maxn];LL solve(){    LL sum = 0;    LL ans = 0;    LL n;    Clean(f,0);    int len = strlen(str);    int cnt = 0;    char last = str[0];    n = 1;    rep(i,1,len-1)    {        if ( str[i] != last )        {            cnt++;            f[cnt] = n * n;            num[cnt] = n;            sum+=f[cnt];            n = 1;            last = str[i];        }        else n++;    }    cnt++;    num[cnt] = n;    f[cnt] = n * n;    sum+=f[cnt];    ans = sum;    rep(i,1,cnt-1)    {        if ( num[i] == 1 && i != 1 )            ans = max( ans , sum - f[i] - f[i-1] - f[i+1] + ( num[i+1] + num[i-1] + 1 )*( num[i+1] + num[i-1] + 1 ) );        else            ans = max( ans , sum-f[i]-f[i+1] + max(  (num[i]+1)*(num[i]+1) + (num[i+1]-1)*(num[i+1]-1) , (num[i]-1)*(num[i]-1) + (num[i+1]+1)*(num[i+1]+1) ) );    }    return ans;}int main(){    cin>>T;    getchar();    rep(kase,1,T)    {        gets(str);        printf("Case #%d: %I64d\n",kase,solve());    }    return 0;}

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