B. Blown Garland-干嘛写dfs过这题啊，明明只要一维数组就可以了

Codeforces Round #392 (Div. 2) B. Blown Garland
题目链接
一开始写了递归来解决，虽然过了，但是赛后看大神的代码，大神思路很简单，
因为题目保证输入是正确的，而且R’, ‘B’, ‘Y’ ,’G’至少有一盏，所以肯定遍历一遍’R’, ‘B’, ‘Y’ ,’G’的最后位置就会有了，注意是最后位置，拿着这个最后位置可以每4个一组的去测试原字符串，肯定是符合的，不符合的输入肯定是错的。
先贴长的递归。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <vector>using namespace std;typedef long long int ll;const int maxn = 150;char s[maxn];char temp[maxn];vector<int> v;int r,b,y,g;int len;bool testcan(int plast,char *t,int inde){    bool canR,canB,canY,canG;    canR = canB = canY = canG = true;//  printf("%d ",plast);    if(plast+1<len){                if(t[plast+1]=='R')canR = false;            if(t[plast+1]=='B')canB = false;            if(t[plast+1]=='Y')canY = false;            if(t[plast+1]=='G')canG = false;        }    if(plast+2<len){            if(t[plast+2]=='R')canR = false;            if(t[plast+2]=='B')canB = false;            if(t[plast+2]=='Y')canY = false;            if(t[plast+2]=='G')canG = false;            }    if(plast+3<len){            if(t[plast+3]=='R')canR = false;            if(t[plast+3]=='B')canB = false;            if(t[plast+3]=='Y')canY = false;            if(t[plast+3]=='G')canG = false;            }    if(plast-1>=0){            if(t[plast-1]=='R')canR = false;            if(t[plast-1]=='B')canB = false;            if(t[plast-1]=='Y')canY = false;            if(t[plast-1]=='G')canG = false;        }    if(plast-2>=0){            if(t[plast-2]=='R')canR = false;            if(t[plast-2]=='B')canB = false;            if(t[plast-2]=='Y')canY = false;            if(t[plast-2]=='G')canG = false;        }    if(plast-3>=0){            if(t[plast-3]=='R')canR = false;            if(t[plast-3]=='B')canB = false;            if(t[plast-3]=='Y')canY = false;            if(t[plast-3]=='G')canG = false;        }    if(canR){    //      printf("R\n");        t[plast]='R';        r++;        if( inde - 1  < 0 )return true;        if(testcan(v[inde-1],t,inde-1))return true;        else canR = false;        r--;    }    if(canB){    //      printf("B\n");        t[plast]='B';        b++;        if( inde - 1  < 0 )return true;        if(testcan(v[inde-1],t,inde-1))return true;        else canB = false;        b--;    }    if(canY){    //      printf("Y\n");        t[plast]='Y';        y++;        if( inde - 1  < 0 )return true;        if(testcan(v[inde-1],t,inde-1))return true;        else canY = false;        y--;    }    if(canG){        //  printf("G\n");        t[plast]='G';        g++;        if( inde - 1  < 0 )return true;        if(testcan(v[inde-1],t,inde-1))return true;        else canG = false;        g--;    }    if(canR == false && canY==false && canB==false && canG==false){        t[plast] = s[plast];        return false;    }    return true;}int main(){//  freopen("in.txt","r",stdin);    scanf("%s",s);    len = strlen(s);        for(int i=0;i<len;++i)temp[i]=s[i];    temp[len]='\0';    r = b = y = g =0;    for(int i=0;i<len;++i)    if(s[i]=='!')v.push_back(i);    if(!v.empty())testcan(v[v.size()-1],temp,v.size()-1);       else {        for(int i=0;i<len;++i){            if(s[i]=='!'){                if(temp[i]=='R')r++;                if(temp[i]=='B')b++;                if(temp[i]=='Y')y++;                if(temp[i]=='G')g++;            }        }           }    printf("%d %d %d %d\n",r,b,y,g);    return 0;}

再贴简单的版本，其中ha数组是计数的，a[4]是遍历一遍更新到最后 一定走得通的rygb组合

#include <iostream>#include <cstring>#include <cstdio>using namespace std;typedef long long int ll;const int maxn = 150;char s[maxn],a[4];int ha[maxn]={0};int main(){    scanf("%s",s);    int len = strlen(s);    for(int i=0;i<len;++i)if(s[i]!='!')a[i%4] = s[i];    for(int i=0;i<len;++i)if(s[i]=='!')ha[ a[i%4] ]++;    printf("%d %d %d %d\n",ha['R'],ha['B'],ha['Y'],ha['G']);    return 0;   }