The Happy Worm 分类： POJ 排序 2015-08-03 18:57 5人阅读 评论(0) 收藏

The Happy Worm
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 4698 Accepted: 1047

Description
The Happy Worm lives in an m*n rectangular field. There are k stones placed in certain locations of the field. (Each square of the field is either empty, or contains a stone.) Whenever the worm sleeps, it lies either horizontally or vertically, and stretches so that its length increases as much as possible. The worm will not go in a square with a stone or out of the field. The happy worm can not be shorter than 2 squares.

The question you are to answer is how many different positions this worm could be in while sleeping.

Input
The first line of the input contains a single integer t (1 <= t <= 11), the number of test cases, followed by the input data for each test case. The first line of each test case contains three integers m, n, and k (1 <= m,n,k <= 131072). The input for this test case will be followed by k lines. Each line contains two integers which specify the row and column of a stone. No stone will be given twice.

Output
There should be one line per test case containing the number of positions the happy worm can be in.

Sample Input

1
5 5 6
1 5
2 3
2 4
4 2
4 3
5 1

Sample Output

9
题意大致理解，不过在实现的过程中却又不少的疑问，希望大神们给指点一下。

#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define LL long long
#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define WW freopen("output.txt","w",stdout)
#define RR freopen("input.txt","r",stdin)

using namespace std;

const int MAX=  140010 ;

struct point
{
    int x;
    int y;
}a[MAX];
int sum;
bool cmp1(point a,point b)
{
    if(a.x<b.x||(a.x==b.x&&a.y<b.y))
    {
        return true;
    }
    return false;
}
bool cmp2(point a,point b)
{
    if(a.y<b.y||(a.y==b.y&&a.x<b.x))
    {
        return true;
    }
    return false;
}
int main()
{
   int T;
   int n,m,k;
   scanf("%d",&T);
   while(T--)
   {
       scanf("%d %d %d",&n,&m,&k);
       for(int i=1;i<=k;i++)
       {
           scanf("%d %d",&a[i].x,&a[i].y);
       }
       sum=0;
       a[0].x=1;
       a[0].y=0;
       a[k+1].x=n;
       a[k+1].y=m+1;
       sort(a+1,a+k+1,cmp1);
       for(int i=1;i<=k+1;i++)
       {
           if(a[i].x==a[i-1].x)
           {
               if(a[i].y-a[i-1].y>2)
               {
                   sum++;
               }
           }
           else
           {
               sum+=(a[i].x-a[i-1].x-1);
               if(m-a[i-1].y>=2)//这里为什么是>=,而不是>
               {
                   sum++;
               }
               if(a[i].y-1>=2)
               {
                   sum++;
               }
           }
       }
       a[0].x=0;
       a[0].y=1;
       a[k+1].x=n+1;
       a[k+1].y=m;
       sort(a+1,a+k+1,cmp2);
       for(int i=1;i<=k+1;i++)
       {
           if(a[i].y==a[i-1].y)
           {
               if(a[i].x-a[i-1].x>2)
               {
                   sum++;
               }
           }
           else
           {
               sum+=(a[i].y-a[i-1].y-1);
               if(n-a[i-1].x>=2)//这里为什么是>=,而不是>
               {
                   sum++;
               }
               if(a[i].x-1>=2)
               {
                   sum++;
               }
           }
       }
       printf("%d\n",sum);
   }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。