POJ-1704 Georgia and Bob (阶梯博弈)

布满荆棘的人生 2022-06-10 10:36 180阅读 0赞

##                                 Georgia and Bob       Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, ..., and place N chessmen on different grids, as shown in the following figure for example:    ![c76b52797bb489f0387716c46010da41_v_1502540847][]   Georgia and Bob move the chessmen in turn. Every time a player will choose a chessman, and move it to the left without going over any other chessmen or across the left edge. The player can freely choose number of steps the chessman moves, with the constraint that the chessman must be moved at least ONE step and one grid can at most contains ONE single chessman. The player who cannot make a move loses the game.       Georgia always plays first since "Lady first". Suppose that Georgia and Bob both do their best in the game, i.e., if one of them knows a way to win the game, he or she will be able to carry it out.       Given the initial positions of the n chessmen, can you predict who will finally win the game?      Input  The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case contains two lines. The first line consists of one integer N (1 <= N <= 1000), indicating the number of chessmen. The second line contains N different integers P1, P2 ... Pn (1 <= Pi <= 10000), which are the initial positions of the n chessmen.  Output  For each test case, prints a single line, "Georgia will win", if Georgia will win the game; "Bob will win", if Bob will win the game; otherwise 'Not sure'.  Sample Input  2 3 1 2 3 8 1 5 6 7 9 12 14 17  Sample Output  Bob will win Georgia will win        思路：因为后面的棋子不能越过前面的棋子，最后谁不能移了谁就输，所以这是一道典型的阶梯博弈。我们只需从后往前每两个棋子分一组，组间距离可以不用考虑，组内距离就相当于不同堆上的石子个数，然后按NIM游戏的做法去做就行了。 ##

阶梯博弈详解：http://blog.csdn.net/shahdza/article/details/7858032

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## `#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cstdlib> #include <cmath> #include <vector> #include <queue> #include <map> #include <algorithm> #include <set> #include <functional> using namespace std; typedef long long LL; typedef unsigned long long ULL; const int INF = 1e9 + 5; const int MAXN = 305; const int MOD = 1000000007; const double eps = 1e-8; const double PI = acos(-1.0); int num[1005]; int main() { int T,N; int ans; scanf("%d", &T); while (T--) { scanf("%d", &N); for (int i = 0; i < N; i++) scanf("%d", &num[i]); sort(num, num + N); if (N % 2 == 0) { ans = num[1] - num[0] - 1; for (int i = 3; i < N; i+=2) ans ^= num[i] - num[i - 1] - 1; } else { ans = num[0] - 1; for (int i = 2; i < N; i+=2) ans ^= num[i] - num[i - 1] - 1; } if (ans == 0) printf("Bob will win\n"); else printf("Georgia will win\n"); } }`         ##

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