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\n",ans);
}
return 0;
}