POJ 2127 Greatest Common Increasing Subsequence -- 动态规划

题目地址：http://poj.org/problem?id=2127

Description

You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal possible length.
Sequence S₁ , S₂ , . . . , S_N of length N is called an increasing subsequence of a sequence A₁ , A₂ , . . . , A_M of length M if there exist 1 <= i₁ < i₂ < . . . < i_N <= M such that S_j = A_ij for all 1 <= j <= N , and S_j < S_j+1 for all 1 <= j < N .

Input

Each sequence is described with M --- its length (1 <= M <= 500) and M integer numbers A_i (-2³¹ <= A_i < 2³¹ ) --- the sequence itself.

Output

On the first line of the output file print L --- the length of the greatest common increasing subsequence of both sequences. On the second line print the subsequence itself. If there are several possible answers, output any of them.

Sample Input

5
1 4 2 5 -12
4
-12 1 2 4

Sample Output

2
1 4

状态dp[i][j]表示seq1[i]从1到i与seq2[j]从1到j并以j为结尾的LCIS的长度

状态转移方程：

dp[i][j] = max(dp[i][k]) + 1, if seq1[i] ==seq2[j], 1 <= k  < j

dp[i][j] = dp[i-1][j], if seq1[i] != seq2[j]

#include <stdio.h>
#include <string.h>

#define MAX 501

typedef struct path{
	int x, y;
}Pre;

int seq1[MAX], seq2[MAX];
int len1, len2;
int dp[MAX][MAX]; //状态dp[i][j]记录seq1前i个与seq2前j个并以seq2[j]为结尾的LCIS的长度
Pre pre[MAX][MAX];//pre[i][j]记录前驱
int path[MAX];//根据pre[i][j]回溯可得到LCIS
int index;

int LCIS(){
	int i, j;
	int max, tx, ty;
	int id_x, id_y;
	int tmpx, tmpy;
	//给dp[i][j]、pre[i][j]置初值
	memset(dp, 0, sizeof(dp));
	memset(pre, 0, sizeof(pre));
	for (i = 1; i <= len1; ++i){
		max = 0;
		tx = ty = 0;
		for (j = 1; j <= len2; ++j){
			//状态转移方程
			dp[i][j] = dp[i-1][j];
			pre[i][j].x = i - 1;
			pre[i][j].y = j;
			if (seq1[i] > seq2[j] && max < dp[i-1][j]){
				max = dp[i-1][j];
				tx = i - 1;
				ty = j;
			}
			if (seq1[i] == seq2[j]){
				dp[i][j] = max + 1;
				pre[i][j].x = tx;
				pre[i][j].y = ty;
			}
		}
	}
	//找到LCIS最后的数字的位置
	max = -1;
	for (i = 1; i <= len2; ++i){
		if (dp[len1][i] > max){
			max = dp[len1][i];
			id_y = i;
		}
	}
	id_x = len1;
	index = 0;
	while (dp[id_x][id_y] != 0){
		tmpx = pre[id_x][id_y].x;
		tmpy = pre[id_x][id_y].y;
		//若找到前一对公共点，则添加进路径
		if (dp[tmpx][tmpy] != dp[id_x][id_y]){
			path[index] = seq2[id_y];
			++index;
		}
		id_x = tmpx;
		id_y = tmpy;
	}
	return max;
}

int main(void){
	int i;
	while (scanf("%d", &len1) != EOF){
		for (i = 1; i <= len1; ++i)
			scanf("%d", &seq1[i]);
		scanf("%d", &len2);
		for (i = 1; i <= len2; ++i)
			scanf("%d", &seq2[i]);

		printf("%d
", LCIS());
		--index;
		if (index >= 0)
			printf("%d", path[index]);
		for (i = index - 1; i >= 0; --i){
			printf(" %d", path[i]);
		}
		printf("
");
	}

	return 0;
}