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  • 【HDOJ】4355 Party All the Time

    好久没做过三分的题目了。

      1 /* 4355 */
      2 #include <iostream>
      3 #include <sstream>
      4 #include <string>
      5 #include <map>
      6 #include <queue>
      7 #include <set>
      8 #include <stack>
      9 #include <vector>
     10 #include <deque>
     11 #include <algorithm>
     12 #include <cstdio>
     13 #include <cmath>
     14 #include <ctime>
     15 #include <cstring>
     16 #include <climits>
     17 #include <cctype>
     18 #include <cassert>
     19 #include <functional>
     20 #include <iterator>
     21 #include <iomanip>
     22 using namespace std;
     23 //#pragma comment(linker,"/STACK:102400000,1024000")
     24 
     25 #define sti                set<int>
     26 #define stpii            set<pair<int, int> >
     27 #define mpii            map<int,int>
     28 #define vi                vector<int>
     29 #define pii                pair<int,int>
     30 #define vpii            vector<pair<int,int> >
     31 #define rep(i, a, n)     for (int i=a;i<n;++i)
     32 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
     33 #define clr                clear
     34 #define pb                 push_back
     35 #define mp                 make_pair
     36 #define fir                first
     37 #define sec                second
     38 #define all(x)             (x).begin(),(x).end()
     39 #define SZ(x)             ((int)(x).size())
     40 #define lson            l, mid, rt<<1
     41 #define rson            mid+1, r, rt<<1|1
     42 
     43 typedef struct {
     44     double x, w;
     45 } node_t;
     46 
     47 const double eps = 1e-4;
     48 const int maxn = 50005;
     49 node_t nd[maxn];
     50 int n;
     51 
     52 double calc(double x) {
     53     double ret = 0.0;
     54     
     55     rep(i, 0, n)
     56         ret += pow(fabs(x-nd[i].x), 3) * nd[i].w;
     57         
     58     return ret;
     59 }
     60 
     61 void solve() {
     62     double l = 1e7;
     63     double r = -1e7;
     64     double m1, m2;
     65     double tmp1, tmp2;
     66     
     67     rep(i, 0, n) {
     68         l = min(l, nd[i].x);
     69         r = max(r, nd[i].x);
     70     }
     71     
     72     while (fabs(r-l) > eps) {
     73         m1 = l + (r-l)/3;
     74         m2 = r - (r-l)/3;
     75         tmp1 = calc(m1);
     76         tmp2 = calc(m2);
     77         if (tmp1 < tmp2) {
     78             r = m2;
     79         } else if (tmp1 > tmp2) {
     80             l = m1;
     81         } else {
     82             l = r;
     83         }
     84     }
     85     
     86     double ans = calc(l);
     87     printf("%.0lf
    ", ans);
     88 }
     89 
     90 int main() {
     91     ios::sync_with_stdio(false);
     92     #ifndef ONLINE_JUDGE
     93         freopen("data.in", "r", stdin);
     94         freopen("data.out", "w", stdout);
     95     #endif
     96     
     97     int t;
     98     
     99     scanf("%d", &t);
    100     rep(tt, 1, t+1) {
    101         scanf("%d", &n);
    102         rep(i, 0, n)
    103             scanf("%lf %lf", &nd[i].x, &nd[i].w);
    104         printf("Case #%d: ", tt);
    105         solve();
    106     }
    107     
    108     #ifndef ONLINE_JUDGE
    109         printf("time = %d.
    ", (int)clock());
    110     #endif
    111     
    112     return 0;
    113 }
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  • 原文地址:https://www.cnblogs.com/bombe1013/p/5188380.html
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