2019 Multi-University Training Contest 1 - String

2019 杭電多校1 - 1009 String
題目鏈接：hdu 6586

題解：
一位位地構造答案字符串，每次貪心地加能加入的最小的字符 (判斷能否加入只要判斷加入之後原字符串剩下的後綴中的每種字符的數目能否足夠滿足條件)。

代碼：

#include <bits/stdc++.h>
#define LL long long
#define LD long double
#define ULL unsigned long long
#define UI unsigned int
#define PII pair<int,int>
#define MPII(x,y) pair<int,int>{x,y}
#define _for(i,j,k) for(int i=j;i<=k;i++)
#define for_(i,j,k) for(int i=j;i>=k;i--)
#define efor(i,u) for(int i=head[u];i;i=net[i])
#define lowbit(x) (x&-x)
#define ls(x) x<<1
#define rs(x) x<<1|1
#define inf 0x3fffffff
//#pragma comment(linker, "/STACK:10240000000,10240000000")
using namespace std;
const int maxn = 1e5 + 5;
const int M = 1e9 + 7;
inline int mad(int a,int b){return (a+=b)>=M?a-M:a;}
string s;
int cind[27][maxn],csuf[27][maxn],l[27],r[27],k; 
bool ck(int d,int j,int ind){
    int tmp=0;
    _for(i,0,25){
        if(j!=i&&csuf[i][ind]<l[i]) return false;
        if(j==i&&csuf[i][ind]<l[i]-1) return false;
        if(j!=i&&l[i]>0) tmp+=l[i];
        if(j==i&&l[i]>1) tmp+=l[i]-1;
    }
    return k-d>=tmp;
}
int find_up(int ind,int kk){
    int l=1,r=cind[ind][0]+1,mid;
    while(l<r){
        mid=(l+r)>>1;
        if(cind[ind][mid]>kk) r=mid;
        else l=mid+1;
    }
    return r;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    while(cin>>s){
        int len = s.length(),sum=0;
        cin>>k;
        _for(i,0,25){
            cin>>l[i]>>r[i];sum+=r[i];
        }
        _for(i,0,len-1){
            cind[s[i]-'a'][++cind[s[i]-'a'][0]]=i;
        }
        for_(i,len-1,0){
            csuf[s[i]-'a'][i]=1;
            _for(j,0,25){
                csuf[j][i]+=csuf[j][i+1];
            }
        }
        if(ck(0,-1,0)&&k<=sum){
            int ind=-1;
            _for(i,1,k){
                _for(j,0,25){
                    int p=find_up(j,ind);
                    if((p<=cind[j][0]&&r[j]>0&&len-cind[j][p]>k-i&&ck(i,j,cind[j][p]+1))){
                        ind=cind[j][p];r[j]--;cout<<s[ind];l[j]--;
                        break;
                    }
                }
            }
            cout<<"\n";
        }
        else{
            cout<<"-1\n";
        }
        _for(i,0,26) l[i]=r[i]=0;
        _for(i,0,26) _for(j,0,len) cind[i][j]=csuf[i][j]=0;
    }
    return 0;
}