hdu5073 简单枚举+精度处理

其实这题还是挺简单的，因为移动k个星球后，这k个星球的权值就可以变为0，所以只有剩下的本来就是连着的才是最优解，也就是说要动也是动两端的，那么就O(N)枚举一遍动哪些就好了。

我是在杭电oj题目重现的比赛上做这题，因为之前听人说现场赛时有人用n^2的算法蹭过了，所以我不断蹭，蹭了一个小时都没蹭过。。。~！@#￥%……

先贴一份乱七八糟想蹭过的代码

/*
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
typedef long long LL;
const double eps = 1e-9;
int get_int() {
    int res = 0, ch;
    while (!((ch = getchar()) >= '0' && ch <= '9')) {
        if (ch == EOF)
            return 1 << 30;
    }
    res = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9')
        res = res * 10 + (ch - '0');
    return res;
}
//输入整数(包括负整数)，用法int a = get_int2();
int get_int2() {
    int res = 0, ch, flag = 0;
    while (!((ch = getchar()) >= '0' && ch <= '9')) {
        if (ch == '-')
            flag = 1;
        if (ch == EOF)
            return 1 << 30;
    }
    res = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9')
        res = res * 10 + (ch - '0');
    if (flag == 1)
        res = -res;
    return res;
}
const int MAXN = 50100;
int N, K, data[MAXN];
int ndata[MAXN];
LL sum[MAXN];
double ans;

inline double getCenter(int s, int e) {
    LL su = sum[e];
    if (s > 0) {
        su -= sum[s - 1];
    }
    double ret = su / (e - s + 1.0);
    return ret;
}

void comput(int s, int e, double c) {
    double ret = 0;
    for (int i = s; i <= e; i++) {
        ret += (data[i] - c) * (data[i] - c);
        if (ret > ans) {
            return;
        }
    }
    if (ret < ans) {
        ans = ret;
    }
}

double comput(double c) {
    double ret = 0;
    for (int i = 0; i < N; ) {
        ret += (data[i] - c) * (data[i] - c) * ndata[i];
        i += ndata[i];
    }
    return ret;
}

void work() {
    double cen = getCenter(0, N - 1);
//    printf("cen = %f
", cen);
    ans = comput(cen);
    for (int a = K; a >= 0; a--) {
        if (ans < eps) {
            break;
        }
        int e = N + a - K - 1;
        double tmpc = getCenter(a, e);
        comput(a, e, tmpc);
    }
}

void treat() {
    for (int i = 0; i < N; i++) {
        int d = data[i];
        int j = i + 1;
        while (j < N && data[j] == d) {
            j++;
        }
        int num = j - i;
        for (j--; j >= i; j--) {
            ndata[j] = num - j + i;
        }
    }
}

int main() {
    int T = get_int();
    while (T--) {
        N = get_int();
        K = get_int();
        for (int i = 0; i < N; i++) {
            data[i] = get_int2();
        }
        sort(data, data + N);
        treat();
        sum[0] = data[0];
        for (int i = 1; i < N; i++) {
            sum[i] = sum[i - 1] + data[i];
        }
        work();
        printf("%.10lf
", ans);
    }
    return 0;
}

下面是正常做法，其实相对于上面的代码也就只有一处改进，因为上面那份代码求解(xi-x)^2的时候是依次计算累加的，可以通过展开公式，通过预存前n项平方和的方式来计算，把这个计算过程从O(N)变成O(1)，就可以过了。

不过我还是wa了几发，原因是一开始忘了对N==K和N-1==K的情况作特殊处理了，因为我后面的代码这个地方没单独考虑。

  1 /*
  2  * Author    : ben
  3  */
  4 #include <cstdio>
  5 #include <cstdlib>
  6 #include <cstring>
  7 #include <cmath>
  8 #include <ctime>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <queue>
 12 #include <set>
 13 #include <map>
 14 #include <stack>
 15 #include <string>
 16 #include <vector>
 17 #include <deque>
 18 #include <list>
 19 #include <functional>
 20 #include <cctype>
 21 using namespace std;
 22 typedef long long LL;
 23 const double eps = 1e-9;
 24 const int MAXN = 50100;
 25 int N, K;
 26 LL data[MAXN], sum[MAXN], sum2[MAXN];
 27 double ans;
 28 int get_int() {
 29     int res = 0, ch;
 30     while (!((ch = getchar()) >= '0' && ch <= '9')) {
 31         if (ch == EOF)
 32             return 1 << 30;
 33     }
 34     res = ch - '0';
 35     while ((ch = getchar()) >= '0' && ch <= '9')
 36         res = res * 10 + (ch - '0');
 37     return res;
 38 }
 39 
 40 //输入整数(包括负整数)，用法int a = get_int2();
 41 int get_int2() {
 42     int res = 0, ch, flag = 0;
 43     while (!((ch = getchar()) >= '0' && ch <= '9')) {
 44         if (ch == '-')
 45             flag = 1;
 46         if (ch == EOF)
 47             return 1 << 30;
 48     }
 49     res = ch - '0';
 50     while ((ch = getchar()) >= '0' && ch <= '9')
 51         res = res * 10 + (ch - '0');
 52     if (flag == 1)
 53         res = -res;
 54     return res;
 55 }
 56 inline LL getSum(int from, int to) {
 57     LL ret = sum[to];
 58     if (from > 0) {
 59         ret -= sum[from - 1];
 60     }
 61     return ret;
 62 }
 63 
 64 inline LL getSum2(int from, int to) {
 65     LL ret = sum2[to];
 66     if (from > 0) {
 67         ret -= sum2[from - 1];
 68     }
 69     return ret;
 70 }
 71 
 72 inline double getCenter(int s, int e) {
 73     LL su = sum[e];
 74     if (s > 0) {
 75         su -= sum[s - 1];
 76     }
 77     double ret = su / (e - s + 1.0);
 78     return ret;
 79 }
 80 
 81 inline double comput(int s, int e, double c) {
 82     LL s1 = getSum(s, e);
 83     LL s2 = getSum2(s, e);
 84     double ret = s2 + (e - s + 1.0) * c * c - 2.0 * c * s1;
 85     return ret;
 86 }
 87 
 88 void work() {
 89     double cen = getCenter(0, N - 1);
 90     ans = comput(0, N - 1, cen);
 91     for (int a = 0; a <= K; a++) {
 92         int e = N + a - K - 1;
 93         double tmpc = getCenter(a, e);
 94         double ret = comput(a, e, tmpc);
 95         if (ret < ans) {
 96             ans = ret;
 97         }
 98     }
 99 }
100 
101 int main() {
102 #ifndef ONLINE_JUDGE
103     freopen("data.in", "r", stdin);
104 #endif
105     int T= get_int();
106     while (T--) {
107         N = get_int();
108         K = get_int();
109         for (int i = 0; i < N; i++) {
110             data[i] = get_int2();
111         }
112         if (K == N || N - 1 == K) {
113             printf("0
");
114             continue;
115         }
116         sort(data, data + N);
117         sum[0] = data[0];
118         sum2[0] = data[0] * data[0];
119         for (int i = 1; i < N; i++) {
120             sum[i] = sum[i - 1] + data[i];
121             sum2[i] = sum2[i - 1] + data[i] * data[i];
122         }
123         work();
124         printf("%.10lf
", ans);
125     }
126     return 0;
127 }