zoukankan      html  css  js  c++  java
  • HDU 1159 Common Subsequence (LCS)

    题意:给定两行字符串,求最长公共子序列。

    析:dp[i][j] 表示第一串以 i 个结尾和第二个串以 j 个结尾,最长公共子序列,剩下的就简单了。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    //#include <tr1/unordered_map>
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    //using namespace std :: tr1;
    
    typedef long long LL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const LL LNF = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1e3 + 5;
    const LL mod = 10000000000007;
    const int N = 1e6 + 5;
    const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
    const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
    const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline int Min(int a, int b){ return a < b ? a : b; }
    inline int Max(int a, int b){ return a > b ? a : b; }
    inline LL Min(LL a, LL b){ return a < b ? a : b; }
    inline LL Max(LL a, LL b){ return a > b ? a : b; }
    inline bool is_in(int r, int c){
        return r >= 0 && r < n && c >= 0 && c < m;
    }
    int dp[maxn][maxn];
    char s1[maxn], s2[maxn];
    
    int main(){
        while(scanf("%s %s", s1+1, s2+1) == 2){
            n = strlen(s1+1);
            m = strlen(s2+1);
            memset(dp, 0, sizeof dp);
            int ans = 0;
            for(int i = 1; i <= n; ++i)
                for(int j = 1; j <= m; ++j)
                    if(s1[i] == s2[j]) { dp[i][j] = dp[i-1][j-1] + 1;  ans = Max(ans, dp[i][j]); }
                    else {  dp[i][j] = Max(dp[i-1][j], dp[i][j-1]);  ans = Max(ans, dp[i][j]); }
    
            printf("%d
    ", ans);
        }
        return 0;
    }
    
  • 相关阅读:
    Oracle block 格式
    oracle用户解锁
    如何扩大重做日志(redolog)文件的大小
    Oracle重建临时表空间
    ORA-32004: obsolete and/or deprecated parameter(s) specified
    oracle分布式事务总结(转载)
    spring注解 @Scheduled(cron = "0 0 1 * * *")实现定时的执行任务
    IDEA启动Tomcat报错1099 is already in use
    js中const,var,let区别
    mysql笔记
  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/5934969.html
Copyright © 2011-2022 走看看