Codeforces-gym-101020 problem C. Rectangles
题目链接:http://codeforces.com/gym/101020/problem/C
2.0 s
64 MB
standard input
standard output
You have n rectangles, each of which is described by three integers i, j and k. This indicates that the lower-left corner of the rectangle will be located at the point (i, 0) and the upper-right corner of the rectangle will be located at the point (j, k). For example, if you have a description of four rectangles like this: 0 2 3 0 3 1 1 2 2 2 4 4 The area occupied by these rectangles will be as follows:
The total area in this case will be 14. You are given n and n triples (i, j, k), your task is to compute the total area occupied by the rectangles which are described by the n triples.
The first line will contain a single integer T indicates the number of test cases. Each test case will consist of n + 1 lines, the first one will contain the number of rectangles n and the following n lines will be the description of the rectangles. Each description will be a triple (i, j, k) as mentioned above. The Constraints: 10 <= T <= 20 1 <= n <= 100 0 <= i < 100 1 <= j < 100 i != j 1 <= k <= 100
For each test case, print a single integer in a separated line indicates the total area occupied by the rectangles.
1
4
0 2 3
0 3 1
1 2 2
2 4 4
14
题意概括:你有n个矩形,每个矩形由三个整数i,j和k描述。这表示矩形的左下角将位于点(i,0),矩形的右上角将位于点(j,k)。例如,如果您有四个这样的矩形的描述:0 2 3 0 3 1 1 2 2 2 4 4这些矩形占用的区域如下:
在这种情况下,总面积为14.您将获得n和n三元组(i,j,k),您的任务是计算由n个三元组描述的矩形占据的总面积。
输入
第一行将包含单个整数T表示测试用例的数量。每个测试用例由n + 1行组成,第一行包含矩形n的数量,后面的n行将是矩形的描述。如上所述,每个描述将是三(i,j,k)。约束:10 <= T <= 20 1 <= n <= 100 0 <= i <100 1 <= j <100 i!= j 1 <= k <= 100
输出
对于每个测试用例,在单独的行中打印单个整数表示矩形占用的总面积。
解题思路:看似很复杂,会有很多重复的不只如何下手,其实我们只需要简单定义一个数组,记录每一竖的小矩形的面积,每次输入新的矩形时,只需要比较新的矩形在每个小矩形与原来的是否更高了,如果更高了,就该成新的高度,否则不变。具体详见代码:
#include<iostream>
#include<string.h>
typedef long long ll;
const int maxn=1e6+;
int a[];
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
memset(a,,sizeof(a));
cin>>n;
int sum=;
while(n--)
{
int x,y,z;
cin>>x>>y>>z;
for(int i=x;i<y;i++)
{
if(a[i]<z) a[i]=z; //如果横坐标为i的小矩形的高度小于新矩形的高度,则更新
}
}
for(int i=;i<=;i++)
sum+=a[i];
cout<<sum<<endl;
}
return ;
}
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