徐州网络赛B-BE，GE or NE记忆化搜索博弈论

Posted 2021-01-07 wyboooo

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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 $1 ? 3$ 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 $? 1$ .

That is, if there are three options in a selection, the score will be increased by $1$ , decreased by $1$ , or multiplied by $? 1$ . The score before the selection is $8$ . Then selecting option $1$ will make the score become $9$ , and selecting option $2$ will make the score $7$ and select option $3$ to make the score $? 8$ . Note that the score has an upper limit of $100$ and a lower limit of $? 100$ . If the score is $99$ at this time, an option that makes the score $+ 2$ is selected. After that, the score will change to $100$ 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 $k$ , it will enter a good ending; if it is less than or equal to a certain value $l$ , 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 $k$ , $l$ 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 $k$ value, the $l$ value, 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 $n, m, k, l$ （ $1 \leq n \leq 1000$ , $? 100 \leq m \leq 100$ , $? 100 \leq l < k \leq 100$ ）, 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 $n$ lines contains three integers $a, b, c$ （ $a \geq 0$ , $b \geq 0$ , $c = 0$ or $c = 1$ ）,indicates the options that appear in this selection,in which $a = 0$ means there is no option to increase the score in this selection, $a > 0$ means there is an option in this selection to increase the score by $a$ ; $b = 0$ means there is no option to decrease the score in this selection, $b > 0$ means there is an option in this selection to decrease the score by $b$ ; $c = 0$ means there is no option to multiply the score by $? 1$ in this selection , $c = 1$ means there is exactly an option in this selection to multiply the score by $? 1$ . It is guaranteed that $a, b, c$ are not equal to $0$ 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

样例输出1

Good Ending

样例输入2

样例输出2

Bad Ending

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

题意：

两个人玩游戏初始分值是m 进行n次操作

每次操作有三种选项 a,b,c

a不为0表示可以选择加a；b不为0表示可以选择减b； c不为0表示可以选择乘-1

分值的上下界是-100和100

一个人希望分值最后大于k 一个人希望分值最后小于l 求最后分值会落在哪个范围

思路：

其实不太会记忆化搜索本来以为就是dp 但是写了半天写不出来

其实记忆化搜索就是 dfs递归 + dp记录

dfs过程中一旦发现dp数组已经有值了就直接返回这样对dfs进行了剪枝

用dp[id][x]表示在分值为x时进行第id次操作的结果 1表示先手必胜 0表示平手 -1表示先手必败

根据递归返回的结果加上当前的局数来判断当前结果

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <stdio.h>
  4 #include <vector>
  5 #include <cmath>
  6 #include <cstring>
  7 #include <set>
  8 #include <map>
  9 
 10 #define inf 0x3f3f3f3f
 11 using namespace std;
 12 
 13 typedef long long LL;
 14 
 15 int n, m, l, k;
 16 const int maxn = 1005;
 17 int dp[maxn][205];
 18 struct node{
 19     int a, b, c;
 20 }op[maxn];
 21 
 22 int solve(int id, int x)
 23 {
 24     int win, lose, done, tmp;
 25     if(dp[id][x] <= 1){
 26         return dp[id][x];
 27     }
 28     if(id == n + 1){
 29         if(x >= k) return 1;
 30         if(x <= l) return -1;
 31         return 0;
 32     }
 33 
 34     win = lose = done = 0;
 35     if(id % 2){
 36         if(op[id].a != 0){
 37             tmp = solve(id + 1, min(x + op[id].a, 200));
 38             if(tmp == 1){
 39                 win = 1;
 40             }
 41             if(tmp == 0){
 42                 done = 1;
 43             }
 44             if(tmp == -1){
 45                 lose = 1;
 46             }
 47         }
 48         if(op[id].b != 0){
 49             tmp = solve(id + 1, max(0, x - op[id].b));
 50             if(tmp == 1){
 51                 win = 1;
 52             }
 53             if(tmp == 0){
 54                 done = 1;
 55             }
 56             if(tmp == -1){
 57                 lose = 1;
 58             }
 59         }
 60         if(op[id].c != 0){
 61             tmp = solve(id + 1, 200 - x);
 62             if(tmp == 1){
 63                 win = 1;
 64             }
 65             if(tmp == 0){
 66                 done = 1;
 67             }
 68             if(tmp == -1){
 69                 lose = 1;
 70             }
 71         }
 72         if(win == 1){
 73             return dp[id][x] = 1;
 74         }
 75         else if(done == 1){
 76             return dp[id][x] = 0;
 77         }
 78         else{
 79             return dp[id][x] = -1;
 80         }
 81     }
 82     else{
 83         if(op[id].a != 0){
 84             tmp = solve(id + 1, min(x + op[id].a, 200));
 85             if(tmp == 1){
 86                 lose = 1;
 87             }
 88             if(tmp == 0){
 89                 done = 1;
 90             }
 91             if(tmp == -1){
 92                 win = 1;
 93             }
 94         }
 95         if(op[id].b != 0){
 96             tmp = solve(id + 1, max(0, x - op[id].b));
 97             if(tmp == 1){
 98                 lose = 1;
 99             }
100             if(tmp == 0){
101                 done = 1;
102             }
103             if(tmp == -1){
104                 win = 1;
105             }
106         }
107         if(op[id].c != 0){
108             tmp = solve(id + 1, 200 - x);
109             if(tmp == 1){
110                 lose = 1;
111             }
112             if(tmp == 0){
113                 done = 1;
114             }
115             if(tmp == -1){
116                 win = 1;
117             }
118         }
119         if(win == 1){
120             return dp[id][x] = -1;
121         }
122         else if(done == 1){
123             return dp[id][x] = 0;
124         }
125         else{
126             return dp[id][x] = 1;
127         }
128     }
129 
130 }
131 
132 int main()
133 {
134     while(scanf("%d%d%d%d", &n, &m, &k, &l) != EOF){
135         m += 100;
136         l += 100;
137         k += 100;
138         for(int i = 1; i <= n; i++){
139             scanf("%d%d%d", &op[i].a, &op[i].b, &op[i].c);
140         }
141 
142         //memset(dp, 62, sizeof(dp));
143         //cout<<dp[0]<<endl;
144         memset(dp, inf, sizeof(dp));
145         int ans = solve(1, m);
146         if(ans == 1){
147             printf("Good Ending
");
148         }
149         else if(ans == -1){
150             printf("Bad Ending
");
151         }
152         else{
153             printf("Normal Ending
");
154         }
155     }
156     return 0;
157 }