【CodeForces - 1149B】Three Religions（dp查询）

题目链接
题目大意是给三个栈相当于，q个操作，一种是指定加在某个队列一个字符，第二种是删掉某个栈的一个字符。然后每次操作完问是否满足下列情况：
The three religions can coexist in peace if their descriptions form disjoint subsequences of the Word of Universe. More formally, one can paint some of the characters of the Word of Universe in three colors:1 ,2 ,3 , so that each character is painted in at most one color, and the description of the -th religion can be constructed from the Word of Universe by removing all characters that aren’t painted in color .

貌似第一次见这种类型的用dp来更新查询答案的，感觉挺有意思的，感觉又可以延伸好多问题。
代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<set>
#define INF 0x7f7f7f7f
#define maxx 100005
#define mod 1000000007
using namespace std;
typedef long long ll;
int n,q;
char s[maxx];

int _next[maxx][26];
int pos[26];

int f[255][255][255];

int r1[255],len1;
int r2[255],len2;
int r3[255],len3;
int main()
{
    cin>>n>>q;
    scanf("%s",s+1);
    memset(pos,INF,sizeof(pos));
    for(int i=n;i>=0;i--)
    {
        for(int j=0;j<26;j++)
            _next[i][j]=pos[j];
        if(i)pos[s[i]-'a']=i;
    }
    memset(f,INF,sizeof(f));
    f[0][0][0]=0;
    char c[5];
    int x;
    while(q--)
    {
        scanf("%s",c);
        if(c[0]=='+')
        {
            scanf("%d%s",&x,c);
            int now=c[0]-'a';
            if(x==1)
            {
                r1[++len1]=now;
                for(int j=0;j<=len2;j++)
                    for(int k=0;k<=len3;k++)
                    {
                        f[len1][j][k]=INF;
                        if(f[len1-1][j][k]<=n)
                            f[len1][j][k]=_next[f[len1-1][j][k]][now];                        
                        if(j&&f[len1][j-1][k]<=n)
                            f[len1][j][k]=min(f[len1][j][k],_next[f[len1][j-1][k]][r2[j]]);

                        if(k&&f[len1][j][k-1]<=n)
                            f[len1][j][k]=min(f[len1][j][k],_next[f[len1][j][k-1]][r3[k]]);
                    }
            }
            else if(x==2)
            {
                r2[++len2]=now;
                for(int i=0;i<=len1;i++)
                    for(int k=0;k<=len3;k++)
                {
                    f[i][len2][k]=INF;
                    if(f[i][len2-1][k]<=n)
                        f[i][len2][k]=_next[f[i][len2-1][k]][now];
                    if(i&&f[i-1][len2][k]<=n)
                        f[i][len2][k]=min(f[i][len2][k],_next[f[i-1][len2][k]][r1[i]]);
                    if(k&&f[i][len2][k-1]<=n)
                        f[i][len2][k]=min(f[i][len2][k],_next[f[i][len2][k-1]][r3[k]]);
                }
            }
            else
            {
                r3[++len3]=now;
                for(int i=0;i<=len1;i++)
                    for(int j=0;j<=len2;j++)
                {
                    f[i][j][len3]=INF;
                    if(f[i][j][len3-1]<=n)
                        f[i][j][len3]=_next[f[i][j][len3-1]][now];
                    if(i&&f[i-1][j][len3]<=n)
                        f[i][j][len3]=min(f[i][j][len3],_next[f[i-1][j][len3]][r1[i]]);
                    if(j&&f[i][j-1][len3]<=n)
                        f[i][j][len3]=min(f[i][j][len3],_next[f[i][j-1][len3]][r2[j]]);
                }
            }
        }
        else
        {
            scanf("%d",&x);
            if(x==1)len1--;
            else if(x==2)len2--;
            else len3--;
        }
        if(f[len1][len2][len3]<=n)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}