2015ACM-ICPC长春E题(hdu5531)题解

一、题意

No response.T_T

二、思路

分$n$为奇数或者偶数讨论。

如果$n$是奇数，列出不等式组：$r_1+r_2=d_{1},r_2+r_3=d_{2},r_3+r_4=d_{3},\cdots,r_{n-1}+r_n=d_{n-1},r_n+r_1=d_n$，可以解出，$r_1=\frac{d_{1}+d_{3}+d_{5}+\cdots-d_{2}-d_{4}-\cdots}{2}$，然后再用上述式子依次算出$r_2$至$r_n$。最后判断$r_1$至$r_n$中是否存在负数即可。

如果$n$是偶数。如果奇数起点的边之和$s_1=d_1+d_3+d_5+\cdots+d_{n-1}$不等于$s_2=d_2+d_4+d_6+\cdots+d_n$，说明无解，否则，一定可以三分枚举出$r_1$，然后推出其他半径。要注意的是，三分的下界和上界需要处理出来。否则，面积关于$r_1$的二次函数在$[low,high]$区间内不一定只有一个极小值。最后判断是否所有半径都大于$0$。

三、代码

#include<bits/stdc++.h>
using namespace std;
);
struct point {
    double x, y;
} p[];
], ansr[];
int n;

inline bool eq(double x, double y) {
    );
}

inline double dis(point a, point b) {
    return hypot(a.x - b.x, a.y - b.y);
}

double calc(double r1) {
    ansr[] = r1;
    double sum = PI * r1 * r1, nr;
    ; i <= n; ++i) {
        ansr[i] = d[i - ] - ansr[i - ];
        sum += PI * ansr[i] * ansr[i];
    }
    return sum;
}

int main() {
//    freopen("e.in", "r", stdin);
    int T;
    for(scanf("%d", &T); T--;) {
        ;
        scanf("%d", &n);
        ; i <= n; ++i)scanf("%lf %lf", &p[i].x, &p[i].y);
        ; i <= n; ++i)d[i] = dis(p[i], i < n ? p[i + ] : p[]);
        ) {
            ;
            ; i <= n; i++) {
                )fz += d[i];
                else fz -= d[i];
            }
            ansr[] = fz / ;
            ; i <= n; ++i)ansr[i] = d[i - ] - ansr[i - ];
            ] + ansr[n], d[n]))imp = ;
        }
        else {
            , t2 = ;
            ; i <= n; ++i) {
                )t1 += d[i];
                else t2 += d[i];
            }
            ;
            else {
                , high = min(d[], d[n]), lmid, rmid, s1, s2, sum = ;
                ; i <= n; ++i) {
                    )sum += d[i], high = min(high, sum);
                    else sum -= d[i], low = max(low, sum);
                }
                ;
                while(ttt--) {
                    lmid = (low + high) / ;
                    rmid = (lmid + high) / ;
                    s1 = calc(lmid), s2 = calc(rmid);
                    if(s1 < s2)high = rmid;
                    else low = lmid;
                }
                double r1;
                if(calc(low) < calc(high))r1 = low;
                else r1 = high;
                calc(r1);
            }
        }
        ; i <= n; ++i) {
            )imp = ;
        }
        if(imp) puts("IMPOSSIBLE");
        else {
            ;
            ; i <= n; ++i)ans += ansr[i] * ansr[i];
            ans *= PI;
            printf("%.2f\n", ans);
            ; i <= n; ++i)printf("%.2f\n", ansr[i]);
        }
    }
    ;
}