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;}