【BZOJ3877】【JSOI2014】保龄球

【题目链接】

• 点击打开链接
  

【思路要点】

• 随机\(K\)个起始状态，以及交换顺序，进行爬山算法，每次交换两个位置，检查是否能够让方案更优，直到达到一个局部最优解。
• 时间复杂度\(O(K\alpha N^3)\)，其中\(\alpha\)为爬山算法的迭代次数，很小。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 105;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
int n, m, x[MAXN], y[MAXN], p[MAXN], q[MAXN];
int calc() {
	int ans = 0, flg = 0;
	for (int i = 1; i <= m; i++) {
		ans += x[i] + y[i];
		if (flg >= 1) ans += x[i];
		if (flg >= 2) ans += y[i];
		if (x[i] + y[i] == 10) {
			if (x[i] == 10) flg = 2;
			else flg = 1;
		} else flg = 0;
	}
	return ans;
}
int main() {
	read(n);
	for (int i = 1; i <= n; i++)
		read(x[i]), read(y[i]), p[i] = q[i] = i;
	if (x[n] == 10) {
		m = n + 1; p[n] = m;
		read(x[m]), read(y[m]);
	} else m = n;
	int finalans = 0;
	for (int cnt = 100; cnt >= 1; cnt--) {
		random_shuffle(p + 1, p + n + 1);
		random_shuffle(q + 1, q + n + 1);
		static int tx[MAXN], ty[MAXN];
		for (int i = 1; i <= m; i++) {
			tx[i] = x[i];
			ty[i] = y[i];
		}
		for (int i = 1; i <= n; i++) {
			int s = p[i], t = p[q[i]];
			tx[s] = x[t];
			ty[s] = y[t];
		}
		for (int i = 1; i <= m; i++) {
			x[i] = tx[i];
			y[i] = ty[i];
		}
		bool flg = true;
		int ans = calc();
		while (flg) {
			flg = false;
			for (int i = 1; i <= n; i++)
			for (int j = i + 1; j <= n; j++) {
				int s = p[i], t = p[j];
				swap(x[s], x[t]);
				swap(y[s], y[t]);
				int tmp = calc();
				if (tmp > ans) {
					ans = tmp;
					flg = true;
				} else {
					swap(x[s], x[t]);
					swap(y[s], y[t]);
				}
			}
		}
		chkmax(finalans, ans);
	}
	printf("%d\n", finalans);
	return 0;
}