zoukankan      html  css  js  c++  java
  • CodeForces 91A Newspaper Headline

    题目链接:CodeForces - 91A  Newspaper Headline

    官方题解:

    In this problem letters from s1 should be taken greedily: take the left letter from the right of the last used letter, if there is no necessary letter from the right of the right used letter the the search should be started from the beginning of string s1 and the answer should be increased by one. But the brute solution get TL and have complexity O(Ans * |s1|).
    This solution can be optimized using the following way. For every position in s1 let's precalculate positions of the closest letters from the right of it from the alphabet. It can be done by moving from the right to the left ins s1 and remembering the last position of every type of symbol. This solution have complexity O(|s1| * K + |s2|), where K is a size of alphabet.

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    #include<string>
    #include<cmath>
    #define CLR(a,b) memset((a),(b),sizeof((a)))
    using namespace std;
    
    const int N = 10005;
    const int M = 1e6+5;
    
    char s1[N], s2[M];
    int d[N][26];
    bool vis[26];
    
    int main () {
        scanf("%s %s", s1, s2);
        int i, j;
        CLR(d, -1);
        CLR(vis, 0);
        int len1 = strlen(s1);
        int len2 = strlen(s2);
        for(i = 0; i < len1; ++i) {
            vis[ s1[i]-'a' ] = 1;
            for(j = 0; j <= i; ++j) {
                if(d[j][ s1[i]-'a' ] == -1)
                    d[j][ s1[i]-'a' ] = i;
            }
        }
        for(j = 0; j < len2; ++j) {
            if(vis[ s2[j]-'a' ] == 0) {
                printf("-1
    "); return 0;
            }
        }
        int ans = 0;
        for(j = 0; j < len2; ) {
            i = 0;
            while(i < len1 && j < len2 && d[i][ s2[j]-'a' ] != -1) {
                i = d[i][ s2[j]-'a' ] + 1;
                j++;
            }
            ans++;
        }
        printf("%d
    ", ans);
        return 0;
    }
  • 相关阅读:
    jquery对同级的td做radio限制
    "javascript:void(0)"用法
    SQL 插入查询的最大ID 号 进行批量
    Java数字、货币值和百分数等的格式化处理
    PHP 注意问题
    Android Fragment真正意义上的onResume和onPause
    Android_CodeWiki_03
    Android_CodeWiki_02
    Android_CodeWiki_01
    Android 启动APP黑屏解决方案
  • 原文地址:https://www.cnblogs.com/GraceSkyer/p/6670584.html
Copyright © 2011-2022 走看看