In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to constantly face the choice, each time you choose the game will provide 1-31−3 options, the player can only choose one of them. Each option has an effect on a "score" parameter in the game. Some options will increase the score, some options will reduce the score, and some options will change the score to a value multiplied by -1−1 .
That is, if there are three options in a selection, the score will be increased by 11, decreased by 11, or multiplied by -1−1. The score before the selection is 88. Then selecting option 11 will make the score become 99, and selecting option 22 will make the score 77 and select option 33 to make the score -8−8. Note that the score has an upper limit of 100100 and a lower limit of -100−100. If the score is 9999 at this time, an option that makes the score +2+2 is selected. After that, the score will change to 100100 and vice versa .
After all the choices have been made, the score will affect the ending of the game. If the score is greater than or equal to a certain value kk, it will enter a good ending; if it is less than or equal to a certain value ll, it will enter the bad ending; if both conditions are not satisfied, it will enter the normal ending. Now, Koutarou and Sena want to play the good endings and the bad endings respectively. They refused to give up each other and finally decided to use the "one person to make a choice" way to play the game, Koutarou first choose. Now assume that they all know the initial score, the impact of each option, and the kk, ll values, and decide to choose in the way that works best for them. (That is, they will try their best to play the ending they want. If it's impossible, they would rather normal ending than the ending their rival wants.)
Koutarou and Sena are playing very happy, but I believe you have seen through the final ending. Now give you the initial score, the kk value, the llvalue, and the effect of each option on the score. Can you answer the final ending of the game?
Input
The first line contains four integers n,m,k,ln,m,k,l（1le n le 10001≤n≤1000, -100 le m le 100−100≤m≤100 , -100 le l < k le 100−100≤l<k≤100 ）, represents the number of choices, the initial score, the minimum score required to enter a good ending, and the highest score required to enter a bad ending, respectively.
Each of the next nn lines contains three integers a,b,ca,b,c（age 0a≥0 , bge0b≥0 ,c=0c=0 or c=1c=1）,indicates the options that appear in this selection,in which a=0a=0 means there is no option to increase the score in this selection, a>0a>0 means there is an option in this selection to increase the score by aa ; b=0b=0 means there is no option to decrease the score in this selection, b>0b>0 means there is an option in this selection to decrease the score by bb; c=0c=0means there is no option to multiply the score by -1−1 in this selection , c=1c=1 means there is exactly an option in this selection to multiply the score by -1−1. It is guaranteed that a,b,ca,b,care not equal to 00 at the same time.
Output
One line contains the final ending of the game. If it will enter a good ending,print "Good Ending"(without quotes); if it will enter a bad ending,print "Bad Ending"(without quotes);otherwise print "Normal Ending"(without quotes).
样例输入1
3 -8 5 -5
3 1 1
2 0 1
0 2 1
样例输出1
Good Ending
样例输入2
3 0 10 3
0 0 1
0 10 1
0 2 1
样例输出2
Bad Ending
SOLUTION:
博弈论+dp，记忆化搜索。（挺好的题） 定义dp[i][j]为第i个轮次，分数为j时，所能得到的最优分数为dp[i][j]
n个轮次，0~n-1，并且K玩家先手，所以进行搜索时，轮次t为偶数时，求最大值，否则求最小值。 因为分数可能为负数，所以增加100的偏移量，便于存储。
被这句话搞蒙了：If it's impossible, they would rather normal ending than the ending their rival wants.
每个人都去自己最优的情况，就已经满足这句话了
CODE:
#include<bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 1005
int n, m, k, l;
struct node
{
int a,b,c;
node(){}
node(int _a,int _b,int _c):a(_a),b(_b),c(_c){}
}p[maxn];
int dp[maxn][205];//dp[i][j]为第i个轮次，分数为j时，所能得到的最优分数为dp[i][j]
int dfs(int t,int score)
{
if(t>=n) return score-100;
if(dp[t][score]!=INF) return dp[t][score];
if(t%2==0)//K玩家
{
int ans=score-100,tmp=-100;
if(p[t].a>0)
{
int n_ans=min((int)100,ans+p[t].a);
tmp=max(tmp,dfs(t+1,n_ans+100));
}
if(p[t].b>0)
{
int n_ans=max((int)-100,ans-p[t].b);
tmp=max(tmp,dfs(t+1,n_ans+100));
}
if(p[t].c>0)
{
tmp=max(tmp,dfs(t+1,-ans+100));
}
dp[t][score]=tmp;
return tmp;
}
else//S玩家
{
int ans=score-100,tmp=100;
if(p[t].a>0)
{
int n_ans=min((int)100,ans+p[t].a);
tmp=min(tmp,dfs(t+1,n_ans+100));
}
if(p[t].b>0)
{
int n_ans=max((int)-100,ans-p[t].b);
tmp=min(tmp,dfs(t+1,n_ans+100));
}
if(p[t].c>0)
{
tmp=min(tmp,dfs(t+1,-ans+100));
}
dp[t][score]=tmp;
return tmp;
}
}
Int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin>>n>>m>>k>>l;
for(int i=0;i<maxn;i++) for(int j=0;j<205;j++) dp[i][j]=INF;
for(int i=0;i<n;i++) cin>>p[i].a>>p[i].b>>p[i].c;
int ans=dfs(0,m+100);
if(ans>=k) puts("Good Ending");
else if(ans<=l) puts("Bad Ending");
else puts("Normal Ending");
cin.get(),cin.get();
return 0;
}