思路: 记忆化搜索
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N =1005;
int down,up;
int s;
int n;
int a[N],b[N],c[N];
int dp[N][305];
int high,low;
map<int ,int >id;
int dfs(int pos,int now)
{
//cout<<"pos "<<pos<<" now "<<now-lim<<endl;
if(pos==n+1){
if(now>=high) return 2;
else if(now>low) return 1;
return 0;
}
if(dp[pos][id[now]]!=-1) return dp[pos][id[now]];
if(pos%2==1){
int op=0;
if(a[pos]){
op=max(op,dfs(pos+1,min(now+a[pos],up)));
}
if(b[pos]){
op=max(op,dfs(pos+1,max(now+b[pos],down)));
}
if(c[pos]){
op=max(op,dfs(pos+1,-now));
}
return dp[pos][id[now]]=op;
}
else{
int op=2;
if(a[pos]){
op=min(op,dfs(pos+1,min(now+a[pos],up)));
}
if(b[pos]){
op=min(op,dfs(pos+1,max(now+b[pos],down)));
}
if(c[pos]){
op=min(op,dfs(pos+1,-now));
}
return dp[pos][id[now]]=op;
}
}
int main()
{
int tot=0;
for(int i=-100;i<=100;i++) id[i]=++tot;
down=-100;
up=100;
scanf("%d %d %d %d",&n,&s,&high,&low);
for(int i=1;i<=n;i++){
scanf("%d %d %d",&a[i],&b[i],&c[i]);
b[i]=-b[i];
}
memset(dp,-1,sizeof(dp));
int op=dfs(1,s);
if(op==2){
printf("Good Ending\n");
}
else if(op==1){
printf("Normal Ending\n");
}
else{
printf("Bad Ending\n");
}
return 0;
}