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\n", 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\n", 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\n");
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\n", ans);
125 }
126 return 0;
127 }
时间： 2024-10-12 04:01:59