矩形嵌套

时间限制：3000 ms | 内存限制：65535 KB

难度：4

描述: 有n个矩形，每个矩形可以用a,b来描述，表示长和宽。矩形X(a,b)可以嵌套在矩形Y(c,d)中当且仅当 a<c,b<d或者b<c,a<d（相当于旋转X90度）。例如（1,5）可以嵌套在（6,2）内，但不能嵌套在（3,4）中。你的任务是选出尽可能多的矩形排成一行，使得除最后一个外，每一个矩形都可以嵌套在下一个矩形内。

输入

第一行是一个正正数N(0<N<10)，表示测试数据组数，
每组测试数据的第一行是一个正正数n，表示该组测试数据中含有矩形的个数(n<=1000)
随后的n行，每行有两个数a,b(0<a,b<100)，表示矩形的长和宽

输出

每组测试数据都输出一个数，表示最多符合条件的矩形数目，每组输出占一行

样例输入

样例输出

//C v0.1 AC
#include<stdio.h>
#include <stdlib.h>
#include <string.h>
#include <alloca.h>
struct rectangle {
    int length;
    int width;
} rctg[1000];

int dplis(struct rectangle rctg[], int n);
//order by length asc ,or order by width asc
int compare(const void *a, const void *b)
{
    struct rectangle *rct1 = (struct rectangle *)a;
    struct rectangle *rct2 = (struct rectangle *)b;
    if (rct1->length == rct2->length)
        return (rct1->width - rct2->width);
    return (rct1->length - rct2->length);
}

int main(void)
{
    int i;
    int n;

    scanf("%d", &n);
    getchar();

    while (n--) {
        int m;
        scanf("%d", &m);
        getchar();

        memset(&rctg, 0x00, sizeof(rctg));

        for (i = 0; i < m; i++) {
            scanf("%d%d", &rctg[i].length, &rctg[i].width);
            if (rctg[i].length < rctg[i].width) {
                int tmp;
                tmp = rctg[i].length;
                rctg[i].length = rctg[i].width;
                rctg[i].width = tmp;
            }
        }

        qsort(rctg, m, sizeof(rctg[0]), compare);

        int rctgmax = dplis(rctg, m);

        printf("%d
", rctgmax);
    }
    return 0;
}

int dplis(struct rectangle rctg[], int n)
{
    int *d = (int *)alloca(sizeof(int) * (n));
    int len = 1;
    int i;
    int j;
    for (i = 0; i < n; ++i) {
        d[i] = 1;
        for (j = 0; j < i; ++j) {
            if (rctg[j].width < rctg[i].width && rctg[j].length < rctg[i].length && d[j] + 1 > d[i]) {
                d[i] = d[j] + 1;
            }
        }

        if (d[i] > len) {
            len = d[i];
        }

    }

    return len;
}

View Code

//C v0.1 WA！
#include<stdio.h>
#include <stdlib.h>
#include <string.h>
struct rectangle {
    int length;
    int width;
} rctg[1000];


//order by length asc ,or order by width asc
int compare(const void *a, const void *b)
{
    struct rectangle *rct1 = (struct rectangle *)a;
    struct rectangle *rct2 = (struct rectangle *)b;
    if (rct1->length == rct2->length)
        return (rct1->width - rct2->width);
    return (rct1->length - rct2->length);
}

int main(void)
{
    int i;
    int n;

    scanf("%d", &n);
    getchar();

    while (n--) {
        int m;
        scanf("%d", &m);
        getchar();

        memset(&rctg, 0x00, sizeof(rctg));

        for (i = 0; i < m; i++) {
            scanf("%d%d", &rctg[i].length, &rctg[i].width);
            if (rctg[i].length < rctg[i].width) {
                int tmp;
                tmp = rctg[i].length;
                rctg[i].length = rctg[i].width;
                rctg[i].width = tmp;
            }
        }

        qsort(rctg, m, sizeof(rctg[0]), compare);
        printf("after asc sort
");

        for (i = 1; i < m; i++) {
            printf("%d %d
", rctg[i].length, rctg[i].width);
        }

        int sum = 1;
        int idx = 0;
        for (i = 1; i < m; i++) {
            //printf("%d %d
", rctg[i].length, rctg[i].width);
            if (rctg[idx].length < rctg[i].length && rctg[idx].width < rctg[i].width) {
                ++sum;
                idx = i;
            } else {
                continue;
            }
        }

        printf("%d
", sum);
    }
    return 0;
}

View Code