CodeForces - 31E TV Game【DP】

题目链接：https://codeforces.com/contest/31/problem/E

思路来自洛谷@Binary_Search_Tree

dp[i][j]表示第i位到第n位中A选了j位的最大A+B

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <climits>
#include <cstring>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <map>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
using namespace std;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define lowbit(x) (x & (-x))
#define CASET int _; scanf("%d", &_); for(int kase=1;kase<=_;kase++)

typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;

static const int INF=0x3f3f3f3f;
static const ll INFL=0x3f3f3f3f3f3f3f3f;
static const db EPS=1e-10;
static const db PI=acos(-1.0);
static const int MOD=1e9+7;

template <typename T>
inline void read(T &f) {
    f = 0; T fu = 1; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') { fu = -1; } c = getchar(); }
    while (c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}
template <typename T>
void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}
static const int MAXN=18*2+5;
int n;
char s[MAXN];
ull dp[MAXN][MAXN];
ull p[MAXN];
void dfs(int i,int j)
{
    if(i>2*n) return;
    if(j && dp[i+1][j-1]+p[j-1]*(s[i]-'0')==dp[i][j])
    {
        putchar('H');
        dfs(i+1,j-1);
    }
    else
    {
        putchar('M');
        dfs(i+1,j);
    }
}
int main()
{
    scanf("%d%s",&n,s+1);
    p[0]=1;
    for(int i=1;i<=18;i++) p[i]=p[i-1]*10;
    ull ans=0;
    for(int i=2*n;i>=1;i--)
    {
        for(int j=0;j<=2*n-i+1;j++)
        {
            int k=2*n-i+1-j;
            if(j) dp[i][j]=max(dp[i][j],dp[i+1][j-1]+p[j-1]*(s[i]-'0'));
            if(k) dp[i][j]=max(dp[i][j],dp[i+1][j]+p[k-1]*(s[i]-'0'));
        }
    }
    dfs(1,n);
    return 0;
}

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