Codeforces 31E TV Game 中途相遇法 状压dp

题目链接：点击打开链接

题意：

给定2*n长的数字。

把这个数字拆成2个长度为n的数字，且相对位置不变。使得拆后得到的2个数字的和最大。

输出一个方案。

显然是中途相遇法，先计算左半段，再计算右半段

分别状压左半段和右半段，注意左半段状压后要在末尾补上0。

代码估计哪里有小越界==，数组开大了一点才过。。具体就不查了。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<string>
#include<queue>
#include<math.h>
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x <0) {
		putchar('-');
		x = -x;
	}
	if (x>9) pt(x / 10);
	putchar(x % 10 + '0');
}
using namespace std;
typedef long long ll;
const int N = 18;
string s;
int n;
int a[N << 1], siz[1 << N];
ll ten[N + 10];
ll dp[2][1 << N], d[2][1 << N], sum[2][N+10], status[2][N+10];
void go(int cur, int off){
	dp[cur][0] = 0;
	for (int i = 1; i < (1 << n); i++)
	for (int j = 0; j < n; j++)
	if ((i&(1 << j))>0){
		dp[cur][i] = dp[cur][i ^ (1 << j)] * 10 + a[n-1-j + off];
		break;
	}

	d[cur][(1 << n) - 1] = 0;
	for (int i = (1 << n) - 2; i >= 0; i--)
	for (int j = 0; j < n; j++)
	if ((i&(1 << j)) == 0){
		d[cur][i] = d[cur][i ^ (1 << j)] * 10 + a[n-1-j + off];
		break;
	}
	if (cur == 0){
		for (int i = 0; i < (1 << n); i++){
			dp[cur][i] *= ten[n - siz[i]];
			d[cur][i] *= ten[siz[i]];
		}
	}
	for (int i = 0; i <= n; i++)sum[cur][i] = -1;
	for (int i = 0; i < (1 << n); i++)
		if (dp[cur][i] + d[cur][i] > sum[cur][siz[i]])
		{
			sum[cur][siz[i]] = dp[cur][i] + d[cur][i];
			status[cur][siz[i]] = i;
		}
}
void put(ll x){
	for (int i = n - 1; i >= 0; i--)
	if (x&(1LL << i))putchar('M');
	else putchar('H');
}
int main(){
	ten[0] = 1; for (int i = 1; i < N; i++)ten[i] = ten[i - 1] * 10;
	rd(n); cin >> s;
	for (int i = 0; i < (1 << n); i++){
		int j = i; siz[i] = 0;
		while (j>0){ siz[i]++; j &= j - 1; }
	}
	for (int i = 0; i < s.length(); i++)a[i] = s[i] - '0';
	go(0, 0);
	go(1, n);
	ll ans = -1;
	pair<ll, ll> p;
	for (int i = 0; i <= n; i++)
		if(ans<sum[0][i] + sum[1][n - i])
		{
			ans = sum[0][i] + sum[1][n - i];
			p = pair<ll,ll>(status[0][i], status[1][ n - i ]);
		}
	put(p.first); put(p.second); puts("");
	return 0;
}

Codeforces 31E TV Game 中途相遇法 状压dp

原文：http://blog.csdn.net/qq574857122/article/details/44516993