## hdu1115 Lifting the Stone（几何，求多边形重心模板题卡塔尔

``````#include
#include
#include
using namespace std;
struct point
{
double x,y;
}PP[1000047];
point bcenter(point pnt[],int n){
point p,s;
double tp,area = 0, tpx=0, tpy=0;
p.x=pnt[0].x;
p.y=pnt[0].y;
for(int i=1;i<=n;++i){
s.x=pnt[(i==n)?0:i].x;
s.y=pnt[(i==n)?0:i].y;
tp=(p.x*s.y-s.x*p.y);
area += tp/2;
tpx += (p.x + s.x) * tp;
tpy += (p.y + s.y) * tp;
p.x = s.x; p.y = s.y;
}
s.x=tpx / (6*area);s.y=tpy/(6*area);
return s;
}
int main()
{
int N,t;
scanf("%d",&t);
while(t--){
scanf("%d",&N);
for(int i=0;i
``````

Lifting the

# Lifting the Stone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 5370 Accepted Submission(s): 2239

Problem Description

There are many secret openings in the floor which are covered by a big
heavy stone. When the stone is lifted up, a special mechanism detects
this and activates poisoned arrows that are shot near the opening. The
only possibility is to lift the stone very slowly and carefully. The ACM
team must connect a rope to the stone and then lift it using a pulley.
Moreover, the stone must be lifted all at once; no side can rise before
another. So it is very important to find the centre of gravity and
connect the rope exactly to that point. The stone has a polygonal shape
and its height is the same throughout the whole polygonal area. Your
task is to find the centre of gravity for the given polygon.

Input

The input consists of T test cases. The number of them (T) is given on
the first line of the input file. Each test case begins with a line
containing a single integer N (3 <= N <= 1000000) indicating the
number of points that form the polygon. This is followed by N lines,
each containing two integers Xi and Yi (|Xi|, |Yi| <= 20000). These
numbers are the coordinates of the i-th point. When we connect the
points in the given order, we get a polygon. You may assume that the
edges never touch each other (except the neighboring ones) and that they
never cross. The area of the polygon is never zero, i.e. it cannot
collapse into a single line.

Output

Print exactly one line for each test case. The line should contain
exactly two numbers separated by one space. These numbers are the
coordinates of the centre of gravity. Round the coordinates to the
nearest number with exactly two digits after the decimal point (0.005
rounds up to 0.01). Note that the centre of gravity may be outside the
polygon, if its shape is not convex. If there is such a case in the
input data, print the centre anyway.

Sample Input

2

4

5 0

0 5

-5 0

0 -5

4

1 1

11 1

11 11

1 11

Sample Output

0.00 0.00

6.00 6.00

``````#include<stdio.h>
#include<stdlib.h>
/*==================================================*\
| 求多边形重心
| INIT: pnt[]已按顺时针(或逆时针)排好序;
| CALL: res = bcenter(pnt, n);
\*==================================================*/
struct point
{
double x, y;
}pnt[1000005],res;
point bcenter(point pnt[], int n)//重心
{
point p, s;
double tp, area = 0, tpx = 0, tpy = 0;
p.x = pnt[0].x;
p.y = pnt[0].y;
for (int i = 1; i <= n; ++i)
{
// point: 0 ~ n-1
s.x = pnt[(i == n) ? 0 : i].x;
s.y = pnt[(i == n) ? 0 : i].y;
tp = (p.x * s.y - s.x * p.y);//叉乘
area += tp / 2;
tpx += (p.x + s.x) * tp;
tpy += (p.y + s.y) * tp;
p.x = s.x;
p.y = s.y;
}
s.x = tpx / (6 * area);
s.y = tpy / (6 * area);
return s;
}
int main()
{
int T,N,i;
scanf("%d",&T);
while(T--)
{
scanf("%d",&N);
for(i=0;i<N;i++)
{
scanf("%lf%lf",&pnt[i].x,&pnt[i].y);
}
res=bcenter(pnt, N);
printf("%0.2lf %0.2lf\n",res.x,res.y);
}
return 0;
}
``````

（代表第i个点到第j个点的最短间隔卡塔尔 中，最终再从low[i][j]搜索最短的门径输出。
890ms过的。呵呵，很强，大致一贯不如小编更加慢的呐！嘿嘿（倒霉意思卡塔尔国，代码也相当长。

AC：