Codeforces-gym-101020 problem C. Rectangles

题目链接：http://codeforces.com/gym/101020/problem/C

C. Rectangles

time limit per test

2.0 s

memory limit per test

64 MB

input

standard input

output

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.

Input

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

Output

For each test case, print a single integer in a separated line indicates the total area occupied by the rectangles.

Examples

Input

Copy

140 2 30 3 11 2 22 4 4

Output

Copy

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+60;
int a[105];
using namespace std;
int main()
{int t;cin>>t;while(t--){int n;memset(a,0,sizeof(a));cin>>n;int sum=0;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=0;i<=100;i++)sum+=a[i];cout<<sum<<endl;}return 0;}

转载于:https://www.cnblogs.com/zjl192628928/p/9280159.html

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