POJ-1654Area(求多边形面积)

Area
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 20510 Accepted: 5594
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：

给定一个多边形，求多边形面积

直接从原点开始用叉积求有向面积，中间不要取绝对值，最后取绝对值即可在*0.5即可
要主要不要存点，不然会MLE，还有不要开double，会损失精度，用longlong，最后判一下奇偶性即可

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int maxn= 1e6+5;LL ans;int dx[10]={0,-1, 0, 1,-1, 0, 1,-1, 0, 1};int dy[10]={0,-1,-1,-1, 0, 0, 0, 1, 1, 1};char s[maxn];int main(){  int T;  scanf("%d",&T);  while(T--)  {    scanf("%s",s+1);    int n=strlen(s+1);    if(n<=3)    {      printf("0\n");      continue;    }    ans=0;    int x=0,y=0,x1,y1,x2,y2;    x1=x+dx[s[1]-'0'];    y1=y+dy[s[1]-'0'];    for(int i=2;i<n;++i)    {      x2=x1+dx[s[i]-'0'];      y2=y1+dy[s[i]-'0'];      ans+=x1*y2-x2*y1;      x1=x2;      y1=y2;    }    ans+=x1*y2-x2*y1;    if(ans<0)ans=-ans;     if(ans&1)printf("%lld.5\n",ans>>1);    else printf("%lld\n",ans>>1);  }  return 0;}