51Nod 1006 最长公共子序列LCS DP水题

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题目链接：51 Nod 1006
题意: 求A，B的最长公共子序列并输出。
分析：先求出最长公共子序列的长度，然后再根据最长公共子序列的长度逆序求出最长公共子序列。

#include <cmath>#include <queue>#include <vector>#include <cstdio>#include <string>#include <cstring>#include <iomanip>#include <iostream>#include <algorithm>using namespace std;//#pragma comment(linker, "/STACK:1024000000,1024000000")#define FIN             freopen("input.txt","r",stdin)#define FOUT            freopen("output.txt","w",stdout)#define fst             first#define snd             second#define lson            l, mid, rt << 1#define rson            mid + 1, r, rt << 1 | 1typedef __int64  LL;//typedef long long LL;typedef unsigned int uint;typedef pair<int, int> PII;typedef pair<LL, LL> PLL;const int INF = 0x3f3f3f3f;const double eps = 1e-6;const int MAXN = 1000 + 5;const int MAXM = 600 + 5;char A[MAXN], B[MAXN];int dp[MAXN][MAXN];char R[MAXN];int main() {#ifndef ONLINE_JUDGE    FIN;#endif // ONLINE_JUDGE    while (~scanf("%s %s", A, B)) {        memset(dp, 0, sizeof(dp));        int lenA = strlen(A);        int lenB = strlen(B);        for (int i = 0; i < lenA; i++) {            for (int j = 0; j < lenB; j++) {                if (A[i] == B[j]) {                    if (i == 0 || j == 0) dp[i][j] = 1;                    else dp[i][j] = dp[i - 1][j - 1] + 1;                } else {                    if (i == 0 && j == 0) dp[i][j] = 0;                    else if (i == 0) dp[i][j] = dp[i][j - 1];                    else if (j == 0) dp[i][j] = dp[i - 1][j];                    else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);                }            }        }        // printf("%d\n", dp[lenA - 1][lenB - 1]);        int temp = dp[lenA - 1][lenB - 1];        R[temp] = '\0';        for (int i = lenA - 1; i >= 0; i--) {            if (!temp) break;            for (int j = lenB - 1; j >= 0; j--) {                if (A[i] == B[j] && dp[i][j] == temp) {                    R[--temp] = A[i];                    break;                }            }        }        puts(R);    }    return 0;}

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