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

题意:便是给你三个多边行的点的坐标,求此多边形的重头戏。

后生可畏道求多边形重心的沙盘题!

 

#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
斯通(几何,求多边形重心模板题)题意:正是给你二个多边行的点的坐标,求此多边形的主导。
风华正茂道求多边形重心的模板…

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.

计算几何,多边形的重心。 

 

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

 

 

题意:求多边形的着注重,,,,

 

模版题。

#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;
}

 

 

转 

那是风度翩翩道多起源的dijkstra模板题,作者全数做了一中午,交了十一回,过了4遍(呵呵,很疯狂啊!卡塔 尔(阿拉伯语:قطر‎

大旨和面积以至坐标的关联

看来,满载而归。应该算完全驾驭了这种算法。这是第四回过的代码,用的拔尖笨的方法,对各类源点和终极分别调用dij算法。

三角形的重心坐标是极点坐标的平均值。

放在low[i][j]
(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。

对此平常的多边形(包涵一条线段的情景卡塔 尔(阿拉伯语:قطر‎

 

算法生龙活虎:平时切合凸多边形

AC:

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