题目链接

做法：

• 首先我们根据题意知道这两个人玩游戏，每次都是选取对自己最优的策略。
• 然后在数据规模上，n<=1000，完全可以用记忆化搜索。
• 每个人针对每个选项，都做出这一步的最优策略。然后进行模拟即可

AC代码：

#include<bits/stdc++.h>
#define IO          ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x)       push_back(x)
#define sz(x)       (int)(x).size()
#define sc(x)       scanf("%d",&x)
#define pr(x)       printf("%d\n",x)
#define abs(x)      ((x)<0?-(x):x)
#define all(x)      x.begin(),x.end()
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 1e8+5;
const int maxn = 1e4+5;
const int INF = 0x3f3f3f3f;
struct node
{
    int a,b,c;
}val[1005];
int dp[1005][205];
int n,m,k,l;
int solve(int step,int score)
{
    if(step == n+1) return score;
    if(dp[step][score+100]!=INF)
        return dp[step][score+100];
    if(step&1){
        int ans = -100;
        if(val[step].a>0){
            int mn = min(100,score+val[step].a);//
            ans = max(ans,solve(step+1,mn));
        }
        if(val[step].b>0){
            int mx = max(-100,score-val[step].b);//
            ans = max(ans,solve(step+1,mx));
        }
        if(val[step].c>0){
            ans = max(ans,solve(step+1,-score));
        }
        dp[step][score+100] = ans;
        return ans;
    }else{
        int ans = 100;
        if(val[step].a>0){
            int mn = min(100,score+val[step].a);//
            ans = min(ans,solve(step+1,mn));
        }
        if(val[step].b>0){
            int mx = max(-100,score-val[step].b);//
            ans = min(ans,solve(step+1,mx));
        }
        if(val[step].c>0){
            ans = min(ans,solve(step+1,-score));
        }
        dp[step][score+100] = ans;
        return ans;
    }
}
int main()
{
    #ifdef LOCAL_FILE
    freopen("in.txt","r",stdin);
    #endif // LOCAL
    IO;
    memset(dp,INF,sizeof(dp));
    cin>>n>>m>>k>>l;
    for(int i=1;i<=n;i++)
        cin>>val[i].a>>val[i].b>>val[i].c;
    int score = solve(1,m);
    if(score>=k) cout<<"Good Ending"<<endl;
    else if(score<k && score>l) cout<<"Normal Ending"<<endl;
    else cout<<"Bad Ending"<<endl;
    return 0;
}