zoukankan      html  css  js  c++  java
  • Long Long Message POJ

    题意:

      给你两串字符,要你找出在这两串字符中都出现过的最长子串

    解析:

      先用个分隔符将两个字符串连接起来,再用后缀数组求出height数组的值,找出一个height值最大并且i与i-1的sa值分别在两串字符中就好了

    #include <iostream>
    #include <cstdio>
    #include <sstream>
    #include <cstring>
    #include <map>
    #include <cctype>
    #include <set>
    #include <vector>
    #include <stack>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #define rap(i, a, n) for(int i=a; i<=n; i++)
    #define rep(i, a, n) for(int i=a; i<n; i++)
    #define lap(i, a, n) for(int i=n; i>=a; i--)
    #define lep(i, a, n) for(int i=n; i>a; i--)
    #define rd(a) scanf("%d", &a)
    #define rlld(a) scanf("%lld", &a)
    #define rc(a) scanf("%c", &a)
    #define rs(a) scanf("%s", a)
    #define MOD 2018
    #define LL long long
    #define ULL unsigned long long
    #define Pair pair<int, int>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define _  ios_base::sync_with_stdio(0),cin.tie(0)
    //freopen("1.txt", "r", stdin);
    using namespace std;
    const int maxn = 200010, INF = 0x7fffffff;
    int s[maxn];
    int sa[maxn], t[maxn], t2[maxn], c[maxn], n;
    int ran[maxn], height[maxn];
    
    void get_sa(int m)
    {
        int i, *x = t, *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 = 0; 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-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
            if(p >= n) break;
            m = p;
        }
        int k = 0;
        for(i = 0; i < n; i++) ran[sa[i]] = i;
        for(i = 0; i < n; i++)
        {
            if(k) k--;
            int j = sa[ran[i]-1];
            while(s[i+k] == s[j+k]) k++;
            height[ran[i]] = k;
        }
    }
    
    char s1[maxn], s2[maxn];
    int main()
    {
        while(~scanf("%s%s", s1, s2)){
            n = 0;
            int len1 = strlen(s1);
            rep(i, 0, len1)
                s[n++] = s1[i] - 'a' + 1;
    
            s[n++] = 28;
            int len2 = strlen(s2);
            rep(i, 0, len2)
                s[n++] = s2[i] - 'a' + 1;
            s[n++] = 0;
            get_sa(30);
            int maxx = -INF;
            rep(i, 1, n)
            {
                if(height[i] > maxx && sa[i] < len1 && sa[i-1] > len1 && sa[i] >= 0)
                    maxx = height[i];
                else if(height[i] > maxx && sa[i-1] < len1 && sa[i] > len1 && sa[i-1] >= 0)
                    maxx = height[i];
            }
            cout<< maxx <<endl;
        }
    
        return 0;
    }
    自己选择的路,跪着也要走完。朋友们,虽然这个世界日益浮躁起来,只要能够为了当时纯粹的梦想和感动坚持努力下去,不管其它人怎么样,我们也能够保持自己的本色走下去。
  • 相关阅读:
    leetcode 48. Rotate Image
    leetcode 203. Remove Linked List Elements 、83. Remove Duplicates from Sorted List 、82. Remove Duplicates from Sorted List II(剑指offer57 删除链表中重复的结点) 、26/80. Remove Duplicates from Sorted ArrayI、II
    leetcode 263. Ugly Number 、264. Ugly Number II 、313. Super Ugly Number 、204. Count Primes
    leetcode 58. Length of Last Word
    安卓操作的一些问题解决
    leetcode 378. Kth Smallest Element in a Sorted Matrix
    android studio Gradle Build速度加快方法
    禁用gridview,listview回弹或下拉悬停
    Android Studio找不到FragmentActivity类
    安卓获取ListView、GridView等滚动的距离(高度)
  • 原文地址:https://www.cnblogs.com/WTSRUVF/p/9495416.html
Copyright © 2011-2022 走看看