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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#735123#5312. Levenshtein DistanceOthersTL 0ms3660kbC++142.5kb2024-11-11 17:35:202024-11-11 17:35:20

Judging History

你现在查看的是最新测评结果

  • [2024-11-11 17:35:20]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:3660kb
  • [2024-11-11 17:35:20]
  • 提交

answer

#include <bits/stdc++.h>
#define ll long long
#define int long long
#define eps 1e-12
#define db long double
#define ull unsigned long long
#define wr(x,ch) write(x),putchar(ch)
#define lb(x) ((x)&-(x))
#define inf (0x3f3f3f3f3f3f3f3f)
using namespace std;
typedef pair<ll,ll> pai;
#define x first
#define y second
ll read() {
	ll x=0;
	bool f=0;char c=getchar();
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<3)+(x<<1)+(c^48),c=getchar();
	return f?-x:x;
}
void write(ll x) {
	if(x<0) putchar('-'),x=-x;
	if(x>=10) write(x/10);
	putchar(x%10^48);
	return ;
}
const int N=1e5+5,K=65,ba=10086001,mod=1e9+9;
int f[K][K],k,n,m,ans[K],vis[K];
char a[N],b[N];
struct Yanami {
	int base[N],ha[N],hb[N];
	void init() {
		base[0]=1;
		for(int i=1;i<=max(n,m);i++) base[i]=base[i-1]*ba%mod;
		for(int i=1;i<=n;i++) ha[i]=(ha[i-1]*ba+a[i])%mod;
		for(int i=1;i<=m;i++) hb[i]=(hb[i-1]*ba+b[i])%mod;
		return ;
	}
	int ask(int l,int r,int *has) {
		return (has[r]-has[l-1]*base[r-l+1]%mod+mod)%mod;
	}
	int ask(int x,int y) {
		int l=0,r=min(n-x+1,m-y+1),mid;
		while(l<r) {
			mid=l+r+1>>1;
			if(ask(x,x+mid-1,ha)==ask(y,y+mid-1,hb)) l=mid;
			else r=mid-1;
		}
		return l;
	}
}LCP;
void chkmax(int &x,int y) {
	if(y>x) x=y;
	return ;
}
signed main() {
	k=read();
	scanf("%s%s",a+1,b+1);
	n=strlen(a+1),m=strlen(b+1);
	LCP.init();
	for(int l=1;l<=m;l++) {
		for(int i=0;i<=k;i++) {
			for(int j=0;j<=2*k;j++) 
				f[i][j]=-inf,vis[j]=0;
		}
		f[0][k]=LCP.ask(1,l);
		// printf("%d::\n",l);
		for(int i=0;i<=k;i++) {
			for(int j=1;j<=2*k-1;j++) {
				if(f[i][j]<0) continue;
				int x=f[i][j],y=l+f[i][j]+(j-k)-1;
				if(x<n&&y<m) chkmax(f[i+1][j],f[i][j]+LCP.ask(x+2,y+2)+1);
				if(x<n&&y<m-1&&LCP.ask(x+1,y+2)) chkmax(f[i+1][j+1],f[i][j]+LCP.ask(x+1,y+2));
				if(x<n) chkmax(f[i+1][j-1],f[i][j]+LCP.ask(x+2,y+1)+1);
				// if(x==n&&y<m) chkmax(f[i+1][j+1],f[i][j]);
			}
			for(int j=0;j<2*k;j++) {
				if(l+f[i][j]+(j-k)-1<m) 
					chkmax(f[i+1][j+1],f[i][j]);
				chkmax(f[i+1][j-1],f[i][j]);
			}
			for(int j=0;j<=2*k;j++) 
				if(vis[j]) f[i][j]=-inf;
				else if(f[i][j]==n) vis[j]=1;
			// printf("%d : ",i);
			// for(int j=0;j<=2*k;j++) printf("%4d ",f[i][j]==-inf?-1:f[i][j]);puts("");
			for(int j=0;j<=2*k;j++) 
				if(f[i][j]==n&&k-j<n) 
					ans[i]++;
		}
	}
	// wr(ans[k],'\n');
	// puts("----------");
	for(int i=0;i<=k;i++) wr(ans[i],'\n');
	return 0;
}
/*
g++ 1.cpp -o 1.exe -std=c++14 -O2;./1.exe

*/

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3660kb

input:

4
aaa
aabbaab

output:

0
5
15
7
1

result:

ok 5 number(s): "0 5 15 7 1"

Test #2:

score: -100
Time Limit Exceeded

input:

30
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA...

output:


result: