• HDU4622: Reincarnation


    题面

    给定一个串,每次询问(l~r)组成的子串的不同子串个数
    串长(2000)

    Sol

    每个左端点建一个(sam)同时记录答案

    # include <bits/stdc++.h>
    # define IL inline
    # define RG register
    # define Fill(a, b) memset(a, b, sizeof(a))
    # define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
    using namespace std;
    typedef long long ll;
    
    IL int Input(){
    	RG int x = 0, z = 1; RG char c = getchar();
    	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    	return x * z;
    }
    
    const int maxn(4005);
    
    int tot = 1, last = 1, trans[26][maxn], fa[maxn], len[maxn], n, ans[maxn][maxn];
    char s[maxn];
    
    IL void Extend(RG int c){
    	RG int p = last, np = ++tot; last = tot;
    	len[np] = len[p] + 1;
    	while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
    	if(!p) fa[np] = 1;
    	else{
    		RG int q = trans[c][p];
    		if(len[q] == len[p] + 1) fa[np] = q;
    		else{
    			RG int nq = ++tot;
    			len[nq] = len[p] + 1, fa[nq] = fa[q];
    			for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
    			fa[np] = fa[q] = nq;
    			while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
    		}
    	}
    }
    
    int main(RG int argc, RG char* argv[]){
    	for(RG int T = Input(); T; --T){
    		scanf(" %s", s + 1), n = strlen(s + 1);
    		for(RG int l = 1; l <= n; ++l){
    			Fill(fa, 0), Fill(len, 0), Fill(trans, 0), tot = last = 1;
    			for(RG int r = l; r <= n; ++r){
    				Extend(s[r] - 'a' + 1);
    				ans[l][r] = ans[l][r - 1] + len[last] - len[fa[last]];
    			}
    		}
    		for(RG int Q = Input(); Q; --Q){
    			RG int l = Input(), r = Input();
    			printf("%d
    ", ans[l][r]);
    		}
    	}
    	return 0;
    }
    
  • 相关阅读:
    Java进阶知识查漏补缺05
    Java进阶知识查漏补缺04
    Java进阶知识查漏补缺03
    Java进阶知识查漏补缺02
    Java进阶知识查漏补缺01
    在IDEA下导入项目后,WEB网页只显示jsp源码的解决方法
    Tutorial 3_工厂方法模式
    Tutorial 2_简单工厂模式
    Tutorial 1_UML与面向对象程序设计基本原则
    Java知识查漏补缺-04
  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8900464.html
Copyright © 2020-2023  润新知