POJ 2127 Greatest Common Increasing Subsequence(LCIS)

Greatest Common Increasing Subsequence

#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;//记录前驱
            pre[i][j].y = j;//记录前驱

            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;//记录前驱
                pre[i][j].y = ty;//记录前驱
            }
        }
    }
    //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];//找最长的长度
            id_y = i;
        }
    }

    id_x = len1;
    index = 0;
    while (dp[id_x][id_y] != 0) {//回溯找路径
        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]) {//前面的长度小才是我们正确的选择
            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\n", LCIS());
        --index;
        if (index >= 0)
            printf("%d", path[index]);
        for (i = index - 1; i >= 0; --i) {//根据记录的细节，需要从后往前打印
            printf(" %d", path[i]);
        }
        printf("\n");
    }

    return 0;
}