Hdu_1050 & poj_1083 move tables (Greedy)

It is a typical greedy algorithm, but you must pay attention to parity, such:

S: 4 6

E: 5 7

What is the answer to this example? The answer is 20.

The numbers 5 and 6 overlap, so we have to finish the operation twice.

It can be seen from the above that when E is an odd number, if the next S = e + 1, it will need to be moved one more time;

Similarly, when E is an even number, if the next S = E-1, also need to move more time;

Code:

#include <stdio.h>
#include <stdlib.h>
#define N 207
struct node
{
int e;
int s;
}num[N];

int cmp(const void * a, const void *b)
{
return (*(struct node*)a).e - (*(struct node *)b).e;
}

int main()
{
//freopen("data.in", "r", stdin);
    int t, n, i,  j;
    scanf("%d", &t);
while(t--)
    {
        scanf("%d", &n);
for(i = 0; i < n; i++)
        {
            scanf("%d%d", &num[i].s, &num[i].e);
if(num[i].s > num[i].e)      //chang
            {
                num[i].s = num[i].s + num[i].e;
                num[i].e = num[i].s - num[i].e;
                num[i].s = num[i].s - num[i].e;
            }
        }
        qsort(num, n, sizeof(num[0]), cmp);
int sum = 0;
for(i = 0; i < n; i++)
        {
int x = 0;
for(j = i; j < n; j++)
            {
if(num[i].e >= num[j].s)    x++;
else if(num[i].e % 2 == 0)      //"e" is even
                {
if(num[j].s == num[i].e - 1)  
                        x++;
                }
else if(num[i].e % 2)        //"e" is odd
                {
if(num[j].s == num[i].e + 1)
                        x++;
                }
else    break;
            }
if(x > sum)    sum = x;
        }
        printf("%d\n", sum*10);
    }
return 0;
}