HDU5531-Rebuild（平面几何+数学）

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题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=5531

思路

感觉就一数学题，把公式推出来，然后求出最小面积即可，主要是IMPOSSIBLE的判断感觉有点麻烦，IMPOSSIBLE的判断大概有这几点需要注意一下：
1. 区分n是奇数还是偶数
2. 对于每个点的半径r都必须>=0

代码

#include <iostream>#include <cstring>#include <stack>#include <vector>#include <set>#include <map>#include <cmath>#include <queue>#include <sstream>#include <iomanip>#include <fstream>#include <cstdio>#include <cstdlib>#include <climits>#include <deque>#include <bitset>#include <algorithm>using namespace std;#define PI acos(-1.0)#define LL long long#define PII pair<int, int>#define PLL pair<LL, LL>#define mp make_pair#define IN freopen("in.txt", "r", stdin)#define OUT freopen("out.txt", "wb", stdout)#define scan(x) scanf("%d", &x)#define scan2(x, y) scanf("%d%d", &x, &y)#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)#define sqr(x) (x) * (x)#define pr(x) cout << #x << " = " << x << endl#define lc o << 1#define rc o << 1 | 1#define pl() cout << endlconst int maxn = 10000 + 5;const double eps = 1e-6;double x[maxn], y[maxn], l[maxn], a[maxn];int n;double dist(double x1, double y1, double x2, double y2) {    return sqrt(sqr(x1 - x2) + sqr(y1 - y2));}int main() {    int T;    scan(T);    while (T--) {        scan(n);        a[0] = l[0] = x[0] = y[0] = 0.0;        double suma = 0.0;        double suma2 = 0.0;        double flag = 1.0;        double _min = 0.0, _max = 10005.0;        bool tot_flag = false;        for (int i = 1; i <= n; i++) scanf("%lf %lf", &x[i], &y[i]);        for (int i = 1; i <= n; i++) {            if (i != n) l[i] = dist(x[i], y[i], x[i + 1], y[i + 1]);            else l[i] = dist(x[1], y[1], x[n], y[n]);            a[i] = -a[i - 1] + l[i - 1];            if (a[i] < 0 && flag > 0) _min = max(_min, -a[i]);            if (flag < 0 && a[i] > 0) _max = min(_max, a[i]);            if (flag < 0 && a[i] < 0) tot_flag = true;            suma += (flag * a[i]);            suma2 += sqr(a[i]);            flag = -flag;        }        double minr;        if (_min > _max || tot_flag) {            puts("IMPOSSIBLE");            continue;        }        if (!(n & 1)) {            if (a[n] != l[n]) {                puts("IMPOSSIBLE");                continue;            }            minr = -suma / double(n);            if (minr < _min) minr = _min;            else if (minr > _max) minr = _max;        } else {            minr = (l[n] - a[n]) / 2.0;            if (minr < _min || minr > _max) {                puts("IMPOSSIBLE");                continue;            }        }        if (minr < 0) {            puts("IMPOSSIBLE");            continue;        }        double ans = n * sqr(minr) + 2 * suma * minr + suma2;        flag = 1.0;        printf("%.2lf\n", PI * ans);        for (int i = 1; i <= n; i++) {            printf("%.2lf\n", a[i] + flag * minr);            flag = -flag;        }    }    return 0;}

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