POJ 1113--Wall(计算凸包)
Wall Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 40363 Accepted: 13754 Description
Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King’s castle. The King was so greedy, that he would not listen to his Architect’s proposals to build a beautiful brick wall with a perfect shape and nice tall towers. Instead, he ordered to build the wall around the whole castle using the least amount of stone and labor, but demanded that the wall should not come closer to the castle than a certain distance. If the King finds that the Architect has used more resources to build the wall than it was absolutely necessary to satisfy those requirements, then the Architect will loose his head. Moreover, he demanded Architect to introduce at once a plan of the wall listing the exact amount of resources that are needed to build the wall.
Your task is to help poor Architect to save his head, by writing a program that will find the minimum possible length of the wall that he could build around the castle to satisfy King’s requirements.
The task is somewhat simplified by the fact, that the King’s castle has a polygonal shape and is situated on a flat ground. The Architect has already established a Cartesian coordinate system and has precisely measured the coordinates of all castle’s vertices in feet. Input
The first line of the input file contains two integer numbers N and L separated by a space. N (3 <= N <= 1000) is the number of vertices in the King’s castle, and L (1 <= L <= 1000) is the minimal number of feet that King allows for the wall to come close to the castle.
Next N lines describe coordinates of castle’s vertices in a clockwise order. Each line contains two integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that represent the coordinates of ith vertex. All vertices are different and the sides of the castle do not intersect anywhere except for vertices. Output
Write to the output file the single number that represents the minimal possible length of the wall in feet that could be built around the castle to satisfy King’s requirements. You must present the integer number of feet to the King, because the floating numbers are not invented yet. However, you must round the result in such a way, that it is accurate to 8 inches (1 foot is equal to 12 inches), since the King will not tolerate larger error in the estimates. Sample Input
9 100 200 400 300 400 300 300 400 300 400 400 500 400 500 200 350 200 200 200 Sample Output
1628 Hint
结果四舍五入就可以了
- 不是直接求凸包，围住城堡的所需的最小距离，这个是凸包的长度，但是建造的围墙和城堡之间还有一个距离L,所以所求周长比凸包长度要多几段圆弧，所有圆弧的角度和为360°360°360°,所以再加上一个半径为L的圆周长即为所求.
- 步进法
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
const int MAX = 5005;
const double PI = acos(-1.0);
typedef struct point {
double x;
double y;
point() {
}
point(double a, double b) {
x = a;
y = b;
}
point operator -(const point &b)const { //返回减去后的新点
return point(x - b.x, y - b.y);
}
double operator *(const point &b)const { //点乘
return x*b.x + y*b.y;
}
}point;
double dist(point p1, point p2) { //返回平面上两点距离
return sqrt((p1 - p2)*(p1 - p2));
}
int n, c, w, ans[MAX], num, sd[MAX], ta, stk[MAX], tp1, tp2;
point x[MAX];
bool cmp(point a, point b) { return a.y < b.y || a.y == b.y && a.x < b.x; } //先按y，再按x从小排序
bool cmulti(point p1, point p2, point p3) { //判断p1p0和p2p0的关系，<0,p1p0在p2p0的逆时针方向
return ((p3.x - p1.x)*(p2.y - p1.y) - (p2.x - p1.x)*(p3.y - p1.y))<0;
}
void Jarvis() {
ta = num = 0;
sd[ta++] = 0, sd[ta++] = 1;
sort(x, x + n, cmp);
for (int i = 2; i < n; i++) {
while (ta > 1 && !cmulti(x[sd[ta - 1]], x[sd[ta - 2]], x[i]))//不是外侧点则回溯，且不取内部点
ta--;
sd[ta++] = i;
}
for (int j = 0; j < ta; j++) {
ans[num++] = sd[j];
}
ta = 0;
sd[ta++] = n - 1;
sd[ta++] = n - 2;
for (int i = n - 3; i >= 0; i--) {
while (ta > 1 && !cmulti(x[sd[ta - 1]], x[sd[ta - 2]], x[i]))
ta--;
sd[ta++] = i;
}
for (int j = 0; j < ta; j++) {
ans[num++] = sd[j];
}
}
int main() {
scanf("%d%d", &n, &c);
for (int i = 0; i < n; i++) {
scanf("%lf%lf", &x[i].x, &x[i].y);
}
Jarvis();
double res1 = 0;
for (int i = 0; i < num - 1; i++) {
res1 += dist(x[ans[i]], x[ans[i + 1]]);
}
res1 += 2 * PI * c;
printf("%.0f\n", res1);
return 0;
}- 用Gamham-scan重写了一下vo(* ￣ ▽ ￣ *)ov
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
const int MAX = 5005;
const double PI = acos(-1.0);
typedef struct point {
double x;
double y;
point() {
}
point(double a, double b) {
x = a;
y = b;
}
point operator -(const point &b)const { //返回减去后的新点
return point(x - b.x, y - b.y);
}
double operator *(const point &b)const { //点乘
return x*b.x + y*b.y;
}
}point;
double dist(point p1, point p2) { //返回平面上两点距离
return sqrt((p1 - p2)*(p1 - p2));
}
int n, res[MAX]; //ans为凸包点集坐标,n为点的个数,sd为临时坐标。
int top = 1;
point p[MAX]; //x存放凸包点集
bool cmp(point a, point b) {
if (a.y == b.y) return a.x < b.x;
return a.y < b.y;
}
bool multi(point p1, point p2, point p0) { //判断p1p0和p2p0的关系，<0,p1p0在p2p0的逆时针方向,>0,p1p0在p2p0的顺时针方向
return (p1.x - p0.x)*(p2.y - p0.y) >= (p2.x - p0.x)*(p1.y - p0.y);
}
void Graham(int n) {
int i, len; //top模拟栈顶
sort(p, p + n, cmp);
//少于3个点也就没有办法形成凸包
if (n == 0)return; res[0] = 0;
if (n == 1)return; res[1] = 1;
if (n == 2)return; res[2] = 2;
for (i = 2; i < n; i++) {
while (top&&multi(p[i], p[res[top]], p[res[top - 1]])) //如果当前这个点和栈顶两个点构成折线右拐了，就回溯到上一个点
top--; //弹出栈顶
res[++top] = i; //否则将这个点入栈
}
len = top;
res[++top] = n - 2;
for (int i = n - 3; i >= 0; i--) {
while (top!=len&&multi(p[i], p[res[top]], p[res[top - 1]]))
top--;
res[++top] = i;
}
}
int main() {
int c;
scanf("%d%d", &n, &c);
for (int i = 0; i < n; i++) {
scanf("%lf%lf", &p[i].x, &p[i].y);
}
Graham(n);
double res1 = 0;
for (int i = 0; i < top; i++) {
res1 += dist(p[res[i]], p[res[i + 1]]);
}
res1 += 2 * PI * c;
printf("%.0f\n", res1);
return 0;
}