USACO 1.1.4 Broken Necklace

USACO 1.1.4 Broken Necklace

动态规划。 设bl, br,  rl, rr 分别为项链 i 处向左、向右收集到蓝色、红色珠子数。项链是环形的，只要把两个同样的项链放在一起，则转换成线性。

bl[0] = rl[0] = 0
if necklace[i] = 'r'    rl[i]=rl[i-1]+1, bl[i]=0;
if necklace[i] = 'b'   bl[i]=bl[i-1]+1, rl[i] = 0;
if necklace[i] = 'w'   rl[i] = rl[i-1]+1, bl[i]=bl[i-1]+1

result = max( max(bl[i], rl[i] ) + max(br[i+1], rr[i+1]) )   (0<= i <= 2*n -1)

/**/ /*
ID: lorelei3
PROG: beads
LANG: C++
*/

#include  < fstream >
#include  < string >
#include  < iostream >

using   namespace  std;

int  main() {
    string s;
    int max=0,len;
    ifstream in("beads.in");
    ofstream out("beads.out");
    in>>len>>s;
    s+=s;
    len+=len;

    for(int i=0; i<len-1; ++i){
        int leftIndex=i, rightIndex=i+1;
        char leftColor=s[leftIndex], rightColor=s[rightIndex];
        while((leftIndex-1)>=0){
            leftIndex--;
            if(s[leftIndex]=='w')
                continue;
            else{
                if(leftColor=='w'){
                    leftColor=s[leftIndex];
                    continue;
                }else if(leftColor==s[leftIndex]){
                    continue;
                }else{
                    leftIndex++;
                    break;
                }
            }
        }
        while((rightIndex+1)<len){
            rightIndex++;
            if(s[rightIndex]=='w')
                continue;
            else{
                if(rightColor=='w'){
                    rightColor=s[rightIndex];
                    continue;
                }else if(rightColor==s[rightIndex]){
                    continue;
                }else{
                    rightIndex--;
                    break;
                }
            }
        }
        max = max>(rightIndex-leftIndex)? max : (rightIndex-leftIndex);
    }
    if(max+1>len/2)
        out<<len/2<<endl;
    else
        out<<max+1<<endl;
    return 0;
}