hdoj 5583 Kingdom of Black and White

题目链接：Kingdom of Black and White

题目大意：给你一些由0和1组成的字符串，现在你可以变换至多一个字符变成另一个，求最大贡献，贡献的计算方式是连续相同的字符的和的平方之和

题目思路：我们可以把一块当成一个整体来看，我们可以知道的是，在长度不为1的块中，变中间的数字一定不会导致贡献的增加，只有头尾是可以的，然后对于长度为1的块需要特判，我们在判断块的时候枚举每一块的头尾就好了，贡献计算的时候小心一下就好了

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;

ll T;
char str[maxn];

struct node{
ll head,tail,length;
}kuai[maxn];

void init(){
for(ll i = 0;i < maxn;i++) kuai[i].length = kuai[i].head = kuai[i].tail = 0;
}

int main(){
scanf("%lld",&T);
for(ll Case = 1;Case <= T;Case++){
getchar();
memset(str,'\0',sizeof(str));
init();
scanf("%s",str);
ll l = strlen(str);
ll cot = 1,len = 1;
kuai[cot].head = 0;
for(ll i = 1;i < l;i++){
if(str[i] != str[i-1]){
kuai[cot].tail = i-1;
kuai[cot+1].head = i;
kuai[cot].length = len;
len = 1;
cot++;
}
else len++;
}
kuai[cot].tail = l-1;kuai[cot].length = kuai[cot].tail-kuai[cot].head+1;
ll sum = 0,maxx = -1;
for(ll i = 1;i <= cot;i++){
sum += kuai[i].length*kuai[i].length;
}
maxx = sum;

ll res = sum-kuai[1].length*kuai[1].length+(kuai[1].length-1)*(kuai[1].length-1)-kuai[2].length*kuai[2].length+(kuai[2].length+1)*(kuai[2].length+1);
maxx = max(maxx,res);

res = sum-kuai[cot].length*kuai[cot].length+(kuai[cot].length-1)*(kuai[cot].length-1)-kuai[cot-1].length*kuai[cot-1].length+(kuai[cot-1].length+1)*(kuai[cot-1].length+1);
maxx = max(maxx,res);
for(ll i = 2;i <= cot-1;i++){
if(kuai[i].length == 1) {
res = sum-1-kuai[i+1].length*kuai[i+1].length-kuai[i-1].length*kuai[i-1].length+(kuai[i-1].length+kuai[i+1].length+1)*(kuai[i-1].length+kuai[i+1].length+1);
maxx = max(res,maxx);
continue;
}
res = sum-kuai[i].length*kuai[i].length+(kuai[i].length-1)*(kuai[i].length-1)-kuai[i-1].length*kuai[i-1].length+(kuai[i-1].length+1)*(kuai[i-1].length+1);
maxx = max(res,maxx);
res = sum-kuai[i].length*kuai[i].length+(kuai[i].length-1)*(kuai[i].length-1)-kuai[i+1].length*kuai[i+1].length+(kuai[i+1].length+1)*(kuai[i+1].length+1);
maxx = max(res,maxx);
}
printf("Case #%lld: %lld\n",Case,maxx);
}
return 0;
}