HDU 1866 A + B forever!

A + B forever!

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1171    Accepted Submission(s): 268

Problem Description

As always, A + B is the necessary problem of this warming-up contest. But the patterns and contents are different from the previous ones. Now I come up with a new “A + B” problem for you, the top coders of HDU.
As we say, the addition defined between two rectangles is the sum of their area . And you just have to tell me the ultimate area if there are a few rectangles.
Isn’t it a piece of cake for you? Come on! Capture the bright “accepted” for yourself.

 

Input

There come a lot of cases. In each case, there is only a string in one line. There are four integers, such as “(x1,y1,x2,y2)”, describing the coordinates of the rectangle, with two brackets distinguishing other rectangle(s) from the string. There lies a plus symbol between every two rectangles. Blanks separating the integers and the interpunctions are added into the strings arbitrarily. The length of the string doesn’t exceed 500.
0<=x1,x2<=1000,0<=y1,y2<=1000.

 

Output

For each case, you just need to print the area for this “A+B” problem. The results will not exceed the limit of the 32-signed integer.

 

Sample Input

(1,1,2,2)+(3,3,4,4)
(1,1,3,3)+(2,2,4,4)+(5,5,6,6)

 

Sample Output

2
8

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
int a[1005][1005];
char b[505];
int d[6];
int main()
{
    while(gets(b))
    {
    int len=strlen(b);
    memset(a,0,sizeof(a));
    int ans=0;
    for(int i=0;i<len;i++)
    {
        if(b[i]=='(')
        {
            int num=0;
            int cot=0;int p;int mark=0;
            for( p=i+1;cot<4;p++)
            {
                if(isdigit(b[p]))
                {
                    num=num*10+b[p]-'0';
                    mark=1;
                }
                else if(!mark)
                    continue;
                else
                {
                    d[++cot]=num;
                    num=0;
                    mark=0;
                }
            }
            i=p;
            int x1=min(d[1],d[3]);
            int x2=max(d[1],d[3]);
            int y1=min(d[2],d[4]);
            int y2=max(d[2],d[4]);
            for(int j=x1;j<x2;j++)
            {
                for(int k1=y1;k1<y2;k1++)
                {
                    if(!a[j][k1])
                    {
                        a[j][k1]=1;
                        ans++;
                    }
                }
            }
        }
    }
    printf("%d
",ans);
    }
    return 0;
}