D - Area解题报告（熊禾强）

D - Area

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1654

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

4582567256244865

Sample Output

000.52

解题思路：面积想相消法，比如东北指向的线段的两端点和原点柱组成的三角形利用公式x1*y2-x2*y1则此面积为负，若为西北指向，则面积为负，此方法的好处是消去多余计算

面积

 

解题代码：

#include <stdio.h>  #include <string.h>  #include <math.h>  char str[1000005]; int tmp(int x1,int y1,int x2,int y2){return (x1*y2-x2*y1);}int main()  {     int xall[10]={0,-1,0,1,-1,0,1,-1,0,1};     int yall[10]={0,-1,-1,-1,0,0,0,1,1,1};     int t;    scanf("%d",&t);    getchar();    while(t--)     {      gets(str);     int x1,y1,x2,y2;    __int64 area=0;     x1=0;     y1=0;     int len=strlen(str);        for(int i=0;i<len;i++)                       {            int j=str[i]-'0';         x2=xall[str[i]-'0']+x1;         y2=yall[str[i]-'0']+y1;                 area+=tmp(x1,y1,x2,y2);         x1=x2;         y1=y2;        }      if(area<0) area=-area;        if(area%2==0) printf("%I64d\n",area/2);        else  printf("%I64d.5\n",area/2);        }      return 0;  }