Fzu2267 The Bigger the Better

题意:两个字符串,每次选一个字符串,删掉该字符,重复上述操作直到删除,最后问最小删除的最小字典序是多少

题解:每次对比字符,选择最小的字符来删,如果有相等的字符,那么就要看当前的后缀,选择字典序最小的那个后缀,后缀的字典序可以用后缀数组来搞

#include <iostream>
#include <algorithm>
#include <string.h>
#define maxn 200100
using namespace std;
struct SuffixArray {
    int s[maxn];
    int sa[maxn],ra[maxn],height[maxn],t1[maxn],t2[maxn],c[maxn],n;
    void build_sa(int m) {
        int i,*x=t1,*y=t2;
        for(i=0; i<m; i++) c[i]=0;
        for(i=0; i<n; i++) c[x[i]=s[i]]++;
        for(i=1; i<m; i++) c[i]+=c[i-1];
        for(i=n-1; i>=0; i--) sa[--c[x[i]]]=i;
        for(int k=1; k<=n; k<<=1) {
            int p=0;
            for(i=n-k; i<n; i++) y[p++]=i;
            for(i=0; i<n; i++)if(sa[i]>=k) y[p++]=sa[i]-k;
            for(i=0; i<m; i++) c[i]=0;
            for(i=0; i<n; i++) c[x[y[i]]]++;
            for(i=1; i<m; i++) c[i]+=c[i-1];
            for(i=n-1; i>=0; i--) sa[--c[x[y[i]]]] = y[i];
            swap(x,y);
            p=1,x[sa[0]]=0;
            for(i=1; i<n; i++)
                x[sa[i]]= y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k]? p-1:p++;
            if(p>=n) break;
            m=p;
        }
    }
    void build_height() {
        int i,j,k=0;
        for(i=0; i<n; i++)ra[sa[i]]=i;
        for(i=0; i<n; i++) {
            if(k)k--;
            j=sa[ra[i]-1];
            while(s[i+k]==s[j+k])k++;
            height[ra[i]]=k;
        }
    }
}sa;
int main(){
    int T, n, m, ca = 1;
    ios::sync_with_stdio(0);
    cin>>T;
    while(T--){
        //memset(sa.s, 0, sizeof(sa.s));
        cin>>n>>m;
        for(int i=0;i<n;i++) cin>>sa.s[i];
        sa.s[n] = sa.s[n+m+1] = 0;
        for(int i=n+1;i<n+m+1;i++) cin>>sa.s[i];
        sa.n = n+m+2;
        sa.build_sa(15);
        sa.build_height();
        int p1 = 0, p2 = n+1, num = 0;
        cout<<"Case "<<ca++<<": ";
        while(p1<n&&p2<n+m+1){
            if(sa.ra[p1] > sa.ra[p2]) cout<<sa.s[p1++];
            else cout<<sa.s[p2++];
        }
        while(p1<n) cout<<sa.s[p1++];
        while(p2<n+m+1) cout<<sa.s[p2++];
        cout<<endl;
    }
    return 0;
}
/*
10
9 9
1 2 3 4 5 6 7 8 9
2 3 1 4 6 5 7 9 8

10
1 1
9
9

10
9 9
1 2 3 4 5 6 7 8 9
2 3 1 4 6 5 7 9 8
*/