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\n", ans);

    }

    return 0;

}