洛谷P3906 Hoof Paper, Scissor (记忆化搜索)

这道题问的是石头剪刀布的的出题问题

首先不难看出这是个dp题

其次这道题的状态也很好确定，之前输赢与之后无关，确定三个状态：当前位置，当前手势，当前剩余次数，所以对于剪刀，要么出石头+1分用一次机会，要么不用机会然后也不加分

然后dp一下就行了，很简单的一道题

#include <bits/stdc++.h>
#include <bits/extc++.h>
using namespace std;
#define limit (100000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int kase;
int n,k;
int dp[limit][3][22];

int c[limit];
int dfs(int cur, int gest, int tot){
    if(dp[cur][gest][tot])return dp[cur][gest][tot];
    if(cur > n)return 0;
    int ans = dfs(cur + 1, gest, tot);
    if(c[cur] == gest)++ans;
    if(tot > 0){
        int t,p;
        if(gest == 0)t = 1, p = 2;
        else if(gest == 2)t = 0,p = 1;
        else t = 0, p = 2;
        ans = max({ans, dfs(cur + 1, t, tot - 1), dfs(cur + 1, p, tot - 1)});
    }
    return dp[cur][gest][tot] = ans;
}
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    FASTIO
    cin>>n>>k;
    rep(i,1,n){
        char cc;
        cin>>cc;
        if(cc == 'H'){
            c[i] = 0;
        }else if(cc == 'S'){
            c[i] = 1;
        }else{
            c[i] = 2;
        }
    }
    int ans = 0;

    cout<<max({ans,dfs(1,0,k),dfs(1,1,k),dfs(1,2,k)});;
    return 0;
}

标签：洛谷,cur,int,tot,Scissor,P3906,ans,gest,define
来源： https://www.cnblogs.com/tiany7/p/13753847.html