2017 CCPC 秦皇岛 C

Crusaders Quest is an interesting mobile game. A mysterious witch has brought great darkness to the game world, and the only hope for your kingdom is to save the Goddesses so that they can unleash their power to fight against the witch.

In order to save the game world, you need to choose three heroes to fight for victory and use their skills wisely. Nine skill blocks of three different types (three blocks per type) will be presented at the bottom of the screen. If () consecutive blocks are of the same type, you can tap on them and eliminate them, thus triggering the powerful skill they represent. After the elimination, the blocks to their left will be connected with the blocks to their right. Moreover, if consecutive blocks of the same type are eliminated, the powerful skill they unleash will be upgraded to a super skill, which is the most powerful skill of all.

DreamGrid is a newbie in this game, and he wants to trigger the super skill as many times as he can. Given nine skill blocks satisfying the description above, please help DreamGrid calculate the maximum number of times he can trigger the super skill.

Input
There are multiple test cases. The first line of input contains an integer (about 50), indicating the number of test cases. For each test case:

The first line contains a string () consisting of three ‘g’s, three ‘a’s and three ‘o’s, representing the nine skill blocks of three different types. Each type of character represents one type of skill block.

Output
For each test case, output an integer denoting the maximum number of times DreamGrid can trigger the super skill.

Sample Input
7
gggaaaooo
aaoogggoa
googgaaao
agogaooag
goooggaaa
gogogoaaa
gaogaogao
Sample Output
3
3
2
1
3
2
1
Hint
For the first sample test case, DreamGrid can first eliminate “aaa” (one super skill triggered), thus changing the skill blocks to “gggooo”. He can then eliminate “ggg” (another super skill triggered) and finally eliminate “ooo” (a third super skill triggered). So the answer is 3.

For the second sample test case, DreamGrid can first eliminate “ggg” (one super skill triggered), thus changing the skill blocks to “aaoooa”. He can then eliminate “ooo” (another super skill triggered) and finally eliminate “aaa” (a third super skill triggered). So the answer is also 3.

For the third sample test case, DreamGrid can first eliminate “aaa” (one super skill triggered), thus changing the skill blocks to “googgo”. He can then eliminate “oo” to obtain “gggo”, and eliminate “ggg” (another super skill triggered) to obtain “o”. So the answer is 2. It is easy to prove that he cannot trigger the super skill three times under this arrangement of skill blocks.

题意：给你九个字母，其中三个g，三个a，三个o。每次可以选择释放一个字母，或者两个相邻并相同的字母，或者三个相邻并相同的字母（绝招）。问总共可以释放出几个绝招。

分析：可以dfs。也可以寻找规律，两个字母之间不同字母的个数来进行分类讨论。

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;const int maxn = 1e2+10;int l[4],r[4];int num[4];void doit(char s[],char a1,char a2,char a3){    memset(num,0,sizeof num);    memset(l,-1,sizeof l);    memset(r,-1,sizeof r);    int ll=strlen(s);    for(int i=0;i<ll;i++){        if(s[i]==a1&&l[1]==-1) l[1]=i;        if(s[i]==a2&&l[2]==-1) l[2]=i;        if(s[i]==a3&&l[3]==-1) l[3]=i;        if(s[i]==a1) r[1]=i,num[1]++;        if(s[i]==a2) r[2]=i,num[2]++;        if(s[i]==a3) r[3]=i,num[3]++;    }}int ans=1;void dfs(char s[],char a1,char a2,char a3){    int now=0;    doit(s,a1,a2,a3);    if(num[1]==3) now++;    int len=0;    int slen=9;    char ss[11];    for(int i=0;i<l[1];i++){        ss[len++]=s[i];    }    for(int i=r[1]+1;i<slen;i++){        ss[len++]=s[i];    }    ss[len++]=0;    doit(ss,a1,a2,a3);    slen=len;    len=0;    for(int i=0;i<l[2];i++){        ss[len++]=ss[i];    }    for(int i=r[2]+1;i<slen;i++){        ss[len++]=ss[i];    }    ss[len++]=0;    if(num[2]==3) now++;    doit(ss,a1,a2,a3);    if(num[3]==3) now++;    ans=max(ans,now);}int main(){    int T;    scanf("%d",&T);    while(T--){            char s[11];        ans=1;        scanf("%s",s);        dfs(s,'g','o','a');        dfs(s,'g','a','o');        dfs(s,'a','g','o');        dfs(s,'a','o','g');        dfs(s,'o','a','g');        dfs(s,'o','g','a');        printf("%d\n",ans);    }    return 0;}