poj1654 Area （计算几何）

Area

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18800		Accepted: 5160

Description
You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the following vertexes of the polygon
until back to the initial vertex. For each step you may go North, West, South or East with step length of 1 unit, or go Northwest, Northeast, Southwest or Southeast with step length of square root of 2.

For example, this is a legal polygon to be computed and its area is 2.5:



Input
The first line of input is an integer t (1 <= t <= 20), the number of the test polygons. Each of the following lines contains a string composed of digits 1-9 describing how the polygon is formed by walking from the origin.
Here 8, 2, 6 and 4 represent North, South, East and West, while 9, 7, 3 and 1 denote Northeast, Northwest, Southeast and Southwest respectively. Number 5 only appears at the end of the sequence indicating the stop of walking. You may assume that the input
polygon is valid which means that the endpoint is always the start point and the sides of the polygon are not cross to each other.Each line may contain up to 1000000 digits.
Output
For each polygon, print its area on a single line.
Sample Input

4
5
825
6725
6244865

Sample Output

0
0
0.5
2

Source
POJ Monthly--2004.05.15 Liu Rujia@POJ

题意：一个人初始在原点，按照题目所给走法，求最后得到的矩形的面积；1~9分别表示八个方向，5表示停止。

分析：以起点为原点，每走一条路，把起点和终点分别和原点连接构成两个向量，然后用所得向量求出该三角形的面积；最后矩形的面积就是这若干个三角形的面积和。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 1e9;
const int MOD = 1e9+7;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define N 50010
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int dx[10]={0,1,1,1,0,0,0,-1,-1,-1};
int dy[10]={0,-1,0,1,-1,0,1,-1,0,1};
char str[1000000+10];
ll area,x,y,px,py;

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%s",str);
int len = strlen(str);
if(len<3||'5'==str[0])
{
printf("0\n");
continue;
}
area = 0;
x = y = 0;
for(int i=0; i<len-1; i++)
{
px = x+dx[str[i]-'0'];
py = y+dy[str[i]-'0'];
area += (x*py-y*px);
x = px;
y = py;
}
if(area < 0) area = -area;
printf("%lld",area/2);
if(area % 2) printf(".5");
printf("\n");
}
return 0;
}