【动态规划】最长上升子序列

题意:n个信封和1张卡片,求最多满足左信封<右信封的大小,最左边的信封能装入卡片
输入
3 3 3
5 4
12 11
9 8
输出
3
1 3 2
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct Node
{
    int si, sj;
    unsigned short n;
};

Node Susake[5002];

int short dp[5002][5002], b[5002], xx[5002];//path

int comp(Node a, Node b)
{
    return a.sj > b.sj ? 0 : 1;
}

int main(int argc, char *argv[])
{
    int n, si, sj, Max, fi, fj, ei, ej, k;
    while(scanf("%d%d%d", &n, &si, &sj) != EOF)
    {
        memset(Susake, 0, sizeof(Susake));
        memset(dp, 0, sizeof(dp));
        memset(b, 0, sizeof(b));
        memset(xx, 0, sizeof(xx));
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d", &Susake[i].si, &Susake[i].sj);
            Susake[i].n = i;
        }
        sort(Susake + 1, Susake + n + 1, comp);
        Max = 0;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= i; j++)
            {
                if(Susake[i].si > Susake[j].si && Susake[i].sj > Susake[j].sj &&
                   Susake[i].si > si && Susake[j].si > si &&
                   Susake[i].sj > sj && Susake[j].sj > sj)
                    dp[i][j] = b[j] + 1;
                else if(Susake[i].si <= si || Susake[i].sj <= sj)
                    dp[i][j] = 0;
                else
                    dp[i][j] = 1;
                if(b[i] < dp[i][j])
                {
                    b[i] = dp[i][j];
                    xx[i] = j;
                }
                if(Max < dp[i][j])
                {
                    ei = i; ej = j;//ei
                    Max = dp[i][j];
                }
            }
        }
        xx[1] = 1;
        if(Max)
        {
            printf("%d\n", Max);
            k = 1;
            b[k] = Susake[ei].n;
            if(Max == 1)
                printf("%d\n", b[1]);
            else
            {
                while(1)
                {
                    if(dp[ei][ej] == 1) break;
                    ei = ej;
                    ej = xx[ej];
                    k++;
                    b[k] = Susake[ei].n;
                }
                for(int i = k; i >= 2; i--)
                    printf("%d ", b[i]);
                printf("%d\n", b[1]);
            }
        }
        else printf("0\n");
    }
    return 0;
}