Codeforces 242(DIV 2) B题

来源：互联网发布：奥卡姆剃刀知乎编辑：程序博客网时间：2024/06/06 19:45

思路：有两种方法：（1）二分枚举答案，（2）排序，然后直接扫描，这里有个小技巧，就是每个节点都保存距离比它小的节点的人数之和，这样可以在降低第二步处理的时间复杂度，总是间复杂度为O(nlogn)+O(n) = O(nlogn)。两种方法运行时间一样，一开始我是直接二分的，做完后看题目标签是sort，然后感觉排序可做，就写了一下。

二分版：

#include<cmath>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#define MAXN 1010using namespace std;const double eps = 1e-8;typedef struct Point{    int x, y, man;    double dist;    void d(){         dist = sqrt(x*x+y*y);    }}Point;Point P[MAXN];int n, s;int Judge(double len){    int sum = 0;    for(int i = 0;i < n;i ++){        if(P[i].dist + eps < len) sum += P[i].man;        if(sum + s >= 1000000) return 1;    }    return -1;}double binary_Search(double l, double r){    double ll, rr, mid;    ll = l, rr = r;    while(ll < rr+eps){        mid = (ll + rr)/2;        int flag = Judge(mid);        if(flag == 1) rr = mid-eps;        else if(flag == -1) ll = mid+eps;    }    return ll;}int main(){    double maxn;    /* freopen("in.c", "r", stdin); */    /* freopen("out.cpp" , "w", stdout); */    while(~scanf("%d%d", &n, &s)){        int sum = 0;        maxn = eps;        for(int i = 0;i < n;i ++){            scanf("%d%d%d", &P[i].x, &P[i].y, &P[i].man);            P[i].d();            sum += P[i].man;            maxn = max(maxn, P[i].dist);        }        if(sum + s < 1000000) printf("-1\n");        else if(sum + s == 1000000) printf("%.8lf\n", maxn);        else printf("%.8lf\n", binary_Search(eps, maxn));    }    return 0;}

sort版：

#include<cmath>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#define MAXN 1010const double eps = 1e-8;using namespace std;typedef struct Point{    int x, y, man, sum;    double dist;    void d(){        dist = sqrt(x*x+y*y);    }    bool operator < (const Point &A) const{        return dist < A.dist;    }}Point;Point P[MAXN];int main(){    int n, s;    /* freopen("in.c", "r", stdin); */    while(~scanf("%d%d", &n, &s)){        int sum = 0;        double maxn = eps;        for(int i = 0;i < n;i ++){            scanf("%d%d%d", &P[i].x, &P[i].y, &P[i].man);            P[i].d();            maxn = max(maxn, P[i].dist);            sum += P[i].man;            P[i].sum = P[i].man;        }        if(sum + s < 1000000) printf("-1\n");        else if(sum + s == 1000000) printf("%.8lf\n", maxn);        else{            sort(P, P+n);            for(int i = 0;i < n;i ++){                if(P[i].sum + s >= 1000000){                    printf("%.8lf\n", P[i].dist);                    break;                }                P[i+1].sum += P[i].sum;      //注意这点;            }        }    }}

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