UVA12206 Stammering Aliens

思路

可以二分答案+哈希

判断有没有那个长为L的串出现至少m次即可

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
int n,m;
const int base = 131;
unsigned long long hashx[80000],power[80000];
int ans=0,lastans=0;
char s[80000];
void init(){
    memset(hashx,0,sizeof(hashx));
    memset(power,0,sizeof(power));
    ans=0;
    lastans=0;
}
void get_hash(void){
    power[0]=1;
    for(int i=1;i<=n;i++){
        hashx[i]=hashx[i-1]*base+s[i];
        power[i]=power[i-1]*base;
    }    
}
unsigned long long hashs(int l,int r){
    return hashx[r]-hashx[l-1]*power[r-l+1];
}
struct MapNode{
    int times,pos;
    bool operator < (const MapNode &b) const{
        return b.times<times;
    }
};
map<unsigned long long,MapNode> M;
bool check(int x){
    M.clear();
    ans=0;
    for(int i=1;i+x-1<=n;i++){
        M[hashs(i,i+x-1)].times++;
        M[hashs(i,i+x-1)].pos=i;
    }
    bool f=false;
    for(auto it = M.begin();it!=M.end();it++){
        if((*it).second.times>=m){
            f=true;
            ans=max(ans,(*it).second.pos);
        }
    }
    return f;
}
int main(){
    // freopen("test.in","r",stdin);
    // freopen("test.out","w",stdout);
    while(scanf("%d",&m)==1&&m){
        init();
        scanf("%s",s+1);
        n=strlen(s+1);
        get_hash();
        int l=0,r=n,t=0;
        while(l<=r){
            int mid=(l+r)>>1;
            if(check(mid))
                lastans=ans,t=mid,l=mid+1;
            else
                r=mid-1;
        }
        if(t){
            printf("%d %d
",t,lastans-1);
        }
        else{
            printf("none
");
        }
    }
    return 0;
}