zoukankan      html  css  js  c++  java
  • hdu 5745 la vie en rose

    这道题的官方题解是dp,但是可以暴力出来。改天再研究怎么dp。

    暴力的时候,如果计算sum的时候,调用strlen函数会超时,可见这个函数并不是十分的好。以后能不用尽量不用。

     

    La Vie en rose

    Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 861    Accepted Submission(s): 461


    Problem Description
    Professor Zhang would like to solve the multiple pattern matching problem, but he only has only one pattern string p=p1p2...pm. So, he wants to generate as many as possible pattern strings from p using the following method:

    1. select some indices i1,i2,...,ik such that 1i1<i2<...<ik<|p| and |ijij+1|>1 for all 1j<k.
    2. swap pij and pij+1 for all 1jk.

    Now, for a given a string s=s1s2...sn, Professor Zhang wants to find all occurrences of all the generated patterns in s.
     
    Input
    There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

    The first line contains two integers n and m (1n105,1mmin{5000,n}) -- the length of s and p.

    The second line contains the string s and the third line contains the string p. Both the strings consist of only lowercase English letters.
     
    Output
    For each test case, output a binary string of length n. The i-th character is "1" if and only if the substring sisi+1...si+m1 is one of the generated patterns.
     
    Sample Input
    3 4 1 abac a 4 2 aaaa aa 9 3 abcbacacb abc
     
    Sample Output
    1010 1110 100100100
     
    Author
    zimpha
     
    Source
     
    #include<iostream>
    #include<stdio.h>
    #include<algorithm>
    #include<set>
    #include<map>
    #include<queue>
    #include<stack>
    #include<cmath>
    #include<cstring>
    #include<string>
    using namespace std;
    const int maxx=100005;
    int main(){
        int t;
        scanf("%d",&t);
        while(t--){
            int n,m;
            char s[maxx];
            char p[maxx];
            scanf("%d%d",&n,&m);
            scanf("%s",s);
            scanf("%s",p);
            int sump[maxx];
            int sums[maxx];
            sump[0]=sums[0]=0;
            for(int i=1;i<=n;i++){
                sums[i]=s[i-1]-'a'+sums[i-1];
            }
            for(int i=1;i<=m;i++){
                sump[i]=p[i-1]-'a'+sump[i-1];
            }
            int pis=sump[m];
            int shave=sums[m];
            for(int i=m;i<=n;i++){
                shave=sums[i]-sums[i-m];
               // cout<<"shave: "<<shave<<endl;
                if(pis==shave){
                    int sta=i-m;
                    int pos=sta;
                    int flag=1;
                    for(int j=0;pos<i;j++){
                        if(s[pos]==p[j]){
                            pos++;
                        }else{
                            if(pos+1<i&&s[pos+1]==p[j]&&s[pos]==p[j+1]){
                                pos+=2;
                                j++;
                            }else{
                                flag=0;
                                break;
                            }
                        }
                    }
                    if(flag){
                        printf("1");
                    }else{
                        printf("0");
                    }
                }else{
                    printf("0");
                }
            }
            for(int i=m-1;i>0;i--){
                printf("0");
            }
            printf("
    ");
    
        }
        return 0;
    }
    View Code
  • 相关阅读:
    大白痴学习webmagic
    selenium docs
    centos6.5单用户模式拯救系统
    使用Squid做代理服务器,Squid单网卡透明代理配置详解(转)
    Redundant Call to Object.ToString()
    Specify a culture in string conversion explicitly
    Benefits of Cold Showers: 7 Reasons Why Taking Cool Showers Is Good For Your Health
    Why Creating a Meaningful Morning Routine Will Make You More Successful
    git图示所有分支的历史
    Rewriting History with Git Rebase
  • 原文地址:https://www.cnblogs.com/superxuezhazha/p/5697505.html
Copyright © 2011-2022 走看看