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Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8893 Accepted Submission(s): 3578
Problem Description
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2, ..., ik> of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = <a, b, f, c> is a subsequence of X = <a, b, c, f, b, c> with index sequence <1, 2, 4, 6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.
Sample Input
abcfbc abfcab programming contest abcd mnp
Sample Output
4 2 0
Source
Recommend
Ignatius
呢个系DP既一种经典题型,叫LCS(Longest Common Subsequence)最长公共子序列,系允许吾连续,连续又系另外一种鸟。
首先就系要提到一种叫做最优子结构既性质,相当好用正斗既性质。
设X = <x1, x2, ........xm>, Xm-1 = <x1, x2, .........xm-1>
Y = <y1, y2, ........yn>, Yn-1 = <y1, y2, .........yn-1>
Z = <z1, z2, ........zk>, Zk-1 = <z1, z2, ........zk>
Z = <z1, z2, ........zk>, Zk-1 = <z1, z2, ........zk>
另外Z为X,Y既LCS,就可以引出如下性质:,
1、当xm = ym既时候, Zk-1为Xm-1同埋Yn-1既LCS。(因为zk就系xm(ym), 呢种情况Z = Zk-1 + 1)
2、当xm != ym既时候, Z就系 "Xm-1同Y” 或者 "X同Yn-1" 既LCS。(即系两种可能之间既最大值)
根据呢两个性质就可以即刻写出一个递归定义:
两个序列既LCS包含左两个序列前序既然LCS。
姐系甘,你要求XY既LCS,就首先要求出"Xm-1同 Yn-1"、 "Xm-1同Y” 或者 "X同Yn-1"既LCS。
写成递归式就系下面甘样:
设ts[i][j]储存Xi同Yi既LCS长度。
{ ts[i][j] = 0 i = 0 || j = 0 //注意呢度可以解决空集情况
ts[i][j] = { ts[i][j] = ts[i][j] xi = yi
{ ts[i][j] = Max(ts[i][j - 1], ts[i - 1][j]) xi != yi
但系如果字符串好大, 写递归好明显吾得吃,所以用二维数组递推,时间复杂度O(n*m)。
下面系代码:
4226991 | 2011-07-20 15:59:19 | Accepted | 1159 | 62MS | 24556K | 952 B | C++ | 10SGetEternal{(。)(。)}! |
#include <iostream> #include <string> using namespace std; #define MAXF(x, y) (x > y? x: y) string x, y; int lx, ly, **ts; void print(int *arr, int siz) { int i; for (i = 0; i < siz; i++) printf("%6d", arr[i]); putchar('\n'); } void init() { int i; lx = x.length(); ly = y.length(); ts = new int *[lx + 1]; for (i = 0; i < lx + 1; i++) { ts[i] = new int [ly + 1]; memset(ts[i], 0, sizeof(int) * (ly + 1)); #if 0 print(ts[i], ly + 1); #endif } } int dp() { int i, j; #if 0 printf("test = %d\n", MAXF(8, 7)); #endif for (i = 1; i <= lx; i++) for (j = 1; j <= ly; j++) if (x[i - 1] == y[j - 1]) ts[i][j] = ts[i - 1][j - 1] + 1; else ts[i][j] = MAXF(ts[i][j - 1], ts[i - 1][j]); return ts[lx][ly]; } int main() { while (cin >> x >> y) { #if 0 printf("%s %s\n", x.c_str(), y.c_str()); #endif init(); printf("%d\n", dp()); } return 0; }
搞掂{= =+}
发表评论
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HDU 1370 Biorhythms
2011-08-03 10:27 1193Biorhythms Time Limit: 2000/10 ... -
HDU 1075 What Are You Talking About
2011-08-04 11:00 866What Are You Talking About Tim ... -
HDU 1058 Humble Numbers
2011-08-02 15:55 1225Humble Numbers Time Limit: 200 ... -
HDU 2095 find your present (2)
2011-08-02 16:13 818find your present (2) Time Lim ... -
HDU 1022 Train Problem I
2011-08-02 21:00 1015Train Problem I Time Limit: 20 ... -
2142 HDU box
2011-08-02 21:21 764box Time Limit: 3000/1000 MS ( ... -
HDU 2151 Worm
2011-08-01 20:48 849Worm Time Limit: 1000/1000 MS ... -
HDU 2722 Here We Go(relians) Again
2011-08-02 00:06 1027Here We Go(relians) Again Time ... -
HDU 3791 二叉搜索树
2011-08-02 14:26 1209二叉搜索树 Time Limit: 20 ... -
PKU 2352 Stars
2011-07-31 21:47 1028Stars Time Limit: 1000MS ... -
PKU 2774 Long Long Message
2011-07-31 21:26 904Long Long Message Time Li ... -
PKU 2777 Count Color
2011-07-31 21:31 797Count Color Time Limit: 1 ... -
HDU 2098 分拆素数和
2011-07-31 21:08 1064分拆素数和 Time Limit: 1000/1000 MS ... -
ZOJ 3512 Financial Fraud .
2011-07-31 20:49 1285Financial Fraud Time Limit: 3 ... -
HDU 1798 Tell me the area .
2011-07-31 20:47 1124Tell me the area Time Limit: 3 ... -
HDU 2962 Trucking .
2011-07-31 20:46 685Trucking Time Limit: 20000/100 ... -
HDU 1596 find the safest road .
2011-07-31 20:45 605find the safest road Time Limi ... -
HDU 2553 N皇后问题 .
2011-07-31 20:20 705N皇后问题 Time Limit: 2000/1000 MS ... -
HDU 1392 Surround the Trees .
2011-07-31 20:19 799Surround the Trees Time Limit: ... -
HDU 1234 开门人和关门人 .
2011-07-31 20:17 677开门人和关门人 Time Limit: 2000/1000 ...
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