bzoj 1132 [POI2008]Tro 几何

[POI2008]Tro

Time Limit: 20 Sec Memory Limit: 162 MB
Submit: 1796 Solved: 604
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Description

平面上有N个点. 求出所有以这N个点为顶点的三角形的面积和 N<=3000

Input

第一行给出数字N,N在[3,3000] 下面N行给出N个点的坐标,其值在[0,10000]

Output

保留一位小数,误差不超过0.1

Sample Input

5
0 0
1 2
0 2
1 0
1 1

Sample Output

7.0

HINT

枚举起点，然后求出以该点为起点的所有向量，然后求面积就可以了。

 1 #include<cstring>
 2 #include<cmath>
 3 #include<cstdio>
 4 #include<iostream>
 5 #include<algorithm>
 6 
 7 #define N 3007
 8 #define ll long long
 9 using namespace std;
10 inline int read()
11 {
12     int x=0,f=1;char ch=getchar();
13     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
14     while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
15     return x*f;
16 }
17 
18 int n;ll ans;
19 struct Node
20 {
21     int x,y;
22     friend inline ll operator*(Node x,Node y)
23     {
24         return x.x*y.y-x.y*y.x;
25     }
26     friend inline bool operator<(Node x,Node y)
27     {
28         if (x.y==y.y) return x.x<y.x;
29         return x.y<y.y;
30     }
31 }a[N],b[N];
32 bool cmp(Node x,Node y)
33 {
34     return x*y>0;
35 }
36 
37 void solve()
38 {
39     sort(a+1,a+n+1);
40     for (int i=1;i<=n-2;i++)
41     {
42         int tot=0;ll sumx=0,sumy=0;
43         for (int j=i+1;j<=n;j++)
44             b[++tot].x=a[j].x-a[i].x,
45             b[tot].y=a[j].y-a[i].y;
46         sort(b+1,b+tot+1,cmp);
47         for (int j=1;j<=tot;j++)
48             sumx+=b[j].x,
49             sumy+=b[j].y;
50         for (int j=1;j<=tot;j++)
51         {
52             sumx-=b[j].x;
53             sumy-=b[j].y;
54             ans+=(ll)b[j].x*sumy-b[j].y*sumx;
55         }
56     }
57 }
58 int main()
59 {
60     n=read();
61     for (int i=1;i<=n;i++)
62         a[i].x=read(),a[i].y=read();
63     solve();
64     if (ans&1) printf("%lld.5",ans/2);
65     else printf("%lld.0",ans/2);
66 }
67 #undef ll

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原文地址：https://www.cnblogs.com/fengzhiyuan/p/8834983.html