Circum Triangle UVA

题目链接：https://vjudge.net/problem/UVA-11186

题意：给定一个圆的半径，圆心为(0,0)。给定圆上n个点，圆上任意三个点比不共线，任意三点均可以组成三角形，求能组成的所有三角形的面积之和。

思路：直接n^3枚举所有的三角形，对面积求和即可；n^2容斥推理也可以，参考的是这个博客http://blog.csdn.net/shimmer_/article/details/38613079，讲得很清楚。

代码：

N^3复杂度：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-9;const int  maxn = 10000 + 20;const int  maxt = 300 + 10;const int mod = 10;const int dx[] = {1, -1, 0, 0};const int dy[] = {0, 0, -1, 1};const int Dis[] = {-1, 1, -5, 5};const double inf = 0x3f3f3f3f;const int MOD = 1000;const double PI = acos(-1.0);int n, m, k;int R;struct node{    double x, y;    node(double x = 0.0, double y = 0.0) : x(x), y(y) {}};vector<node> v;double change(double q){    return q / 180.0 * PI;}double Len(node n1, node n2){//求两点之间距离    return sqrt((n1.x - n2.x) * (n1.x - n2.x) + (n1.y - n2.y) * (n1.y - n2.y));}double Area(double a, double b, double c){//已知三边求三角形面积    double p = (a + b + c) / 2.0;    return sqrt(p * (p - a) * (p - b) * (p - c));}double dis[maxn][maxn];int main(){    while(~scanf("%d%d", &n, &R) && n && R){        v.clear();        memset(dis, 0, sizeof  dis);        double q;        for(int i = 0; i < n; ++i){            scanf("%lf", &q);            q = change(q);            v.push_back(node((double)R * cos(q), (double)R * sin(q)));        }        for(int i = 0; i < n; ++i){//需要预处理一下，不然在n^3里求会超时            for(int j = 0; j < i; ++j){                dis[i][j] = dis[j][i] = Len(v[i], v[j]);            }        }        double ans = 0.0;        double len1, len2, len3;        for(int i = 0; i < n; ++i){            for(int j = i + 1; j < n; ++j){                for(int k = j + 1; k < n; ++k){                    ans += Area(dis[i][j], dis[i][k], dis[j][k]);                }            }        }        printf("%.0f\n", ans);    }    return 0;}

N^3复杂度：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-9;const int  maxn = 10000 + 20;const int  maxt = 300 + 10;const int mod = 10;const int dx[] = {1, -1, 0, 0};const int dy[] = {0, 0, -1, 1};const int Dis[] = {-1, 1, -5, 5};const double inf = 0x3f3f3f3f;const int MOD = 1000;const double PI = acos(-1.0);int n, m, k;int R;struct node{    double x, y;    node(double x = 0.0, double y = 0.0) : x(x), y(y) {}};vector<node> v;double change(double q){    return q / 180.0 * PI;}double Area(double x1, double y1, double x2, double y2, double x3, double y3){//已知三角形三点坐标求三角形面积    return fabs((x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1)) / 2.0;}int main(){    while(~scanf("%d%d", &n, &R) && n && R){        v.clear();        double q;        for(int i = 0; i < n; ++i){            scanf("%lf", &q);            q = change(q);            v.push_back(node((double)R * cos(q), (double)R * sin(q)));        }        double ans = 0.0;        double len1, len2, len3;        for(int i = 0; i < n; ++i){            for(int j = i + 1; j < n; ++j){                for(int k = j + 1; k < n; ++k){                    ans += Area(v[i].x, v[i].y, v[j].x, v[j].y, v[k].x, v[k].y);                }            }        }        printf("%.0f\n", ans);    }    return 0;}


N^2复杂度：
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-9;const int  maxn = 10000 + 20;const int  maxt = 300 + 10;const int mod = 10;const int dx[] = {1, -1, 0, 0};const int dy[] = {0, 0, -1, 1};const int Dis[] = {-1, 1, -5, 5};const double inf = 0x3f3f3f3f;const int MOD = 1000;const double PI = acos(-1.0);int n, m, k;int R;double change(double q){    return q / 180.0 * PI;}double deg[maxn];int main(){    while(~scanf("%d%d", &n, &R) && n && R){        memset(deg, 0, sizeof  deg);        double q;        for(int i = 0; i < n; ++i){            scanf("%lf", &q);            deg[i] = change(q);        }        sort(deg, deg + n);        double ans = 0;        for(int i = 0; i < n; ++i){            for(int j = i + 1; j < n; ++j){                ans += (n + 2 * i - 2 * j) * sin(deg[j] - deg[i]);//看博客链接里的讲解很清楚            }        }        ans = ans * R * R / 2.0;        printf("%.0f\n", ans);    }    return 0;}