SRM 624 D2L3: GameOfSegments, game theory, SD-Grundy theorem, Nimber, spraguegrundy

SRM 624 D2L3: GameOfSegments, game theory, SD-Grundy theorem, Nimber, spraguegrundy

Question: http://community.topcoder.com/stat? C = problem_statement & pm = 13204 & rd = 15857

This topic needs to use the classic theory in game theory. The following is a summary of relevant theories:

Impartial Game: a fair Game. The two sides are the same except who starts the Game first.

Nim: a classic Impartial Game. Many games can be equivalent to Nim.

Nimber (Grundy numbers): number of game states. Each status should be a unique Nimber during the game's process.

The pair-grundy theorem: For an Impartial Game, its equivalent number of Nimber in the Game is equal to the value of the Mex function in all its successor States.

Mex: for a set of S, if G is the complement set of S, then Mex {S} = min {G }.

The number of Nimbers in the current game State needs to be obtained based on the theorem of SD-Grundy, and then the Nimbers values in all subsequent states need to be obtained. This is particularly useful when there are multiple "sub-Impartial Games" at the same time, and the number of Nimber in each "sub-Game" status is operated differently or differently, the number of Nimber in this status.

The Code is as follows:

#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <iostream>#include <sstream>#include <iomanip>#include <bitset>#include <string>#include <vector>#include <stack>#include <deque>#include <queue>#include <set>#include <map>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <cstring>#include <ctime>#include <climits>using namespace std;#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)typedef pair<int, int> pii;typedef long long llong;typedef pair<llong, llong> pll;#define mkp make_pair#define FOREACH(it, X) for(__typeof((X).begin()) it = (X).begin(); it != (X).end(); ++it)/*************** Program Begin **********************/class GameOfSegments {public:    int winner(int N) {    int Nimbers[1001];    Nimbers[0] = Nimbers[1] = 0;    for (int i = 2; i <= N; i++) {    set <int> options;    for (int j = 0; j <= i - 2; j++) {    options.insert(Nimbers[j] ^ Nimbers[i - j - 2]);    }    int r = 0;    while (options.count(r)) {    ++r;    }    Nimbers[i] = r;    }    return (Nimbers[N] > 0 ? 1 : 2);    }};/************** Program End ************************/

Refer:

Http://www.cnblogs.com/fishball/archive/2013/01/19/2866311.html

Http://www.cnblogs.com/hsqdboke/archive/2012/04/20/2459796.html

Http://www.cnblogs.com/hsqdboke/archive/2012/04/21/2461034.html