题目链接：https://nanti.jisuanke.com/t/31454

题意：两个人玩游戏，轮流操作，每次操作有三个选项，当前值+x，当前值-x，当前值变成相反数。然后还有一个区间【L，R】，若第最后得分大于等于R，第一个人赢，小于等于L 第二个人赢。否则平局。问最后谁获胜。

题解：记忆化搜索。定义dp【i】【j】为第 i 轮等分为 j 的胜负状况。然后看代码即可

#include<bits/stdc++.h>
using namespace std;

int n,L,R,dp[1005][250];

struct res{
	int x,y,z;
}s[1005];


int  Go(int x,int y,int t = 0){
	if(x == 1) y = min(y+t,215);
	else if(x == 2) y = max(y-t,15);
	else y += 2*(115-y);
	return y;
}
int Sech(int id,int x){
	int win,lose,done,temp;
	if(dp[id][x] <= 1) 
		return dp[id][x];
	if(id == n+1){
		if(x >= R) return 1;
		if(x <= L) return -1;
		return 0;
	}
	win = lose = done = 0;
	if(id%2){
		if(s[id].x != 0){
			temp = Sech(id+1,Go(1,x,s[id].x));
			if(temp == 1) win = 1;
			if(temp == 0) done = 1;
			if(temp == -1) lose = 1;
		}
		if(s[id].y != 0){
			temp = Sech(id+1,Go(2,x,s[id].y));
			if(temp == 1) win = 1;
			if(temp == 0) done = 1;
			if(temp == -1) lose = 1; 
		}
		if(s[id].z != 0){
			temp = Sech(id+1,Go(3,x));
			if(temp == 1) win = 1;
			if(temp == 0) done = 1;
			if(temp == -1) lose = 1;
		}
		if(win) return dp[id][x] = 1;
		else if(done) return dp[id][x] = 0;
		else return dp[id][x] = -1; 
	}
	else {
		if(s[id].x != 0){
			temp = Sech(id+1,Go(1,x,s[id].x));
			if(temp == 1) lose = 1;
			if(temp == 0) done = 1;
			if(temp == -1) win = 1;
		}
		if(s[id].y != 0){
			temp = Sech(id+1,Go(2,x,s[id].y));
			if(temp == 1) lose = 1;
			if(temp == 0) done = 1;
			if(temp == -1) win = 1;
		}
		if(s[id].z != 0){
			temp = Sech(id+1,Go(3,x));
			if(temp == 1) lose = 1;
			if(temp == 0) done = 1;
			if(temp == -1) win = 1;
		}
		if(win) return dp[id][x] = -1;
		else if(done) return dp[id][x] = 0;
		else return dp[id][x] = 1;
	}
}


int main(){
	int ans,m,i;
	scanf("%d%d%d%d",&n,&m,&R,&L);
	R += 115;
	L += 115;
	for(int i = 1 ; i <= n ; i ++)
		scanf("%d%d%d",&s[i].x,&s[i].y,&s[i].z);
	memset(dp,62,sizeof(dp));
	ans = Sech(1,m+115);
	if(ans==1)
		printf("Good Ending\n");
	else if(ans==-1)
		printf("Bad Ending\n");
	else
		printf("Normal Ending\n");
	return 0;
}