ACM-ICPC 2017 Asia Urumqi G. The Mountain

All as we know, a mountain is a large landform that stretches above the surrounding land in a limited area. If we as the tourists take a picture of a distant mountain and print it out, the image on the surface of paper will be in the shape of a particular polygon.

From mathematics angle we can describe the range of the mountain in the picture as a list of distinct points, denoted by $(x_1,y_1)(x1,y1) to (x_n,y_n)(xn,yn). The first point is at the original point of the coordinate system and the last point is lying on the xx-axis. All points else have positive y coordinates and incremental xx coordinates. Specifically, all x coordinates satisfy 0 = x_1 < x_2 < x_3 < ... < x_n0=x1<x2<x3<...<xn. All yy coordinates are positive except the first and the last points whose yy coordinates are zeroes.$

The range of the mountain is the polygon whose boundary passes through points $(x_1,y_1)(x1,y1) to (x_n,y_n)(xn,yn) in turn and goes back to the first point. In this problem, your task is to calculate the area of the range of a mountain in the picture.$

Input

The input has several test cases and the first line describes an integer $\le t \le 20)t(1≤t≤20) which is the total number of cases.$

In each case, the first line provides the integer $\le n \le 100)n(1≤n≤100) which is the number of points used to describe the range of a mountain. Following nn lines describe all points and the ii-th line contains two integers x_ixi and y_i (0 \le x_i, y_i \le 1000)yi(0≤xi,yi≤1000) indicating the coordinate of the ii-th point.$

Output

For each test case, output the area in a line with the precision of

样例输入

样例输出

1.000000
125.000000
60.500000

题目来源

ACM-ICPC 2017 Asia Urumqi

//根据题意首尾的两个点都在x轴上，因此分成前后两个三角形和中间若干个梯形即可。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
int n,t;
const int N=120;
struct Node{
    int x,y;
}nod[N];
int main()
{
    scanf("%d",&t);
    double ans;
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d%d",&nod[i].x,&nod[i].y);
        ans=0;
        for(int i=1;i<n;i++){
            ans+=(nod[i-1].y+nod[i].y)*(nod[i].x-nod[i-1].x)/2.0;
        }
        printf("%.6f\n",ans);
    }
    return 0;
}