标签：style blog color io os ar for sp div
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 }
标签：style blog color io os ar for sp div
原文地址：http://www.cnblogs.com/moonbay/p/4044739.html
