【HDOJ6586】String（枚举）

题意：给定一个由小写字母组成的字符串S，要求从中选出一个长度为k的子序列，使得其字典序最小，并且第i个字母在子序列中出现的次数在[l[i],r[i]]之间

n,k<=1e5

思路：大概就是记一下后缀和然后逐位确定，把能想到的界都给卡上

这种题写错的话大概随机数据拍都能拍出来，就是调起来占大量机时并且体感极差就是了

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,int>P;
 #define N  110000
 #define M  151000
 #define fi first
 #define se second
 #define MP make_pair
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const ll MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       ll INF=1e18;
       ll inf=5e13;
       int dx[]={-,,,};
       int dy[]={,,-,};

 int s[N][],nxt[N][],ans[N],l[N],r[N],t[],a[],n,K;
 char ch[N];

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 int isok(int x,int y,int z)
 {
     int t=nxt[x][z];
     //printf("x=%d y=%d z=%d t=%d\n",x,y,z,t);
     if(t==n+) return ;
     //if(K-y>n-t) return 0;
     if(a[z]+>r[z]) return ;
     a[z]++;
     //rep(i,0,3) printf("%d ",a[i]);
     //printf("\n");
     int tmp=;
     rep(i,,) tmp+=min(s[t+][i],r[i]-a[i]);
     //printf("tmp1=%d\n",tmp);
     if(tmp<K-y)
     {
         a[z]--;
         return ;
     }

     rep(i,,)
      if(a[i]+min(s[t+][i],min(n-t,K-y))<l[i])
      {
          a[z]--;
          return ;
      }

     tmp=;
     rep(i,,) tmp+=max(,l[i]-a[i]);
     //printf("tmp2=%d\n",tmp);
     if(tmp>n-t||tmp>K-y)
     {
         a[z]--;
         return ;
     }
     a[z]--;
     return ;
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     while(scanf("%s",ch+)!=EOF)
     {
         n=strlen(ch+),K=read();
         if(K==) break;
         rep(i,,) t[i]=n+;
         rep(i,,) s[n+][i]=;
         per(i,n,)
         {
             int x=;
             if(i) x=ch[i]-'a';
             rep(j,,) nxt[i][j]=t[j];
             t[x]=i;
             rep(j,,) s[i][j]=s[i+][j];
             s[i][x]++;
         }
         rep(i,,) l[i]=read(),r[i]=read();
         int flag=;
         rep(i,,K) ans[i]=;
         rep(i,,) a[i]=;
         int now=;
         rep(i,,K)
         {
             int k=;
             rep(j,,)
              if(isok(now,i,j)){k=j; now=nxt[now][j]; a[j]++; break;}

             if(k==){flag=; break;}
             ans[i]=k;
             //printf("ans%d=%d now=%d\n",i,k,now);
         }
         //printf("flag=%d\n",flag);
         if(flag)
         {
             rep(i,,K) printf("%c",ans[i]+'a');
             printf("\n");
         }
          else printf("-1\n");
     }

     return ;
 }