Description
Two players play a graph coloring game. They make moves in turn, first player moves first. Initially they take some undirected graph. At each move, a player can color an uncolored vertex with either white or black color (each player can use any color, possibly different at different turns). It's not allowed to color two adjacent vertices with the same color. A player that can't move loses. After playing this game for some time, they decided to study it. For a start, they've decided to study very simple kind of graph — a chain. A chain consists of N vertices, v 1, v 2,..., v N, and N-1 edges, connecting v 1 with v 2, v 2 with v 3,..., v N-1 with v N. Given a position in this game, and assuming both players play optimally, who will win?Input
The first line of input contains the integer N, . The second line of input describes the current position. It contains N digits without spaces. i th digit describes the color of vertex v i: 0 — uncolored, 1 — black, 2 — white. No two vertices of the same color are adjacent.Output
On the only line of output, print " FIRST " (without quotes) if the player moving first in that position wins the game, and " SECOND " (without quotes) otherwise.Sample Input
1 #include <algorithm> 2 #include <map> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <vector> 6 #include <iostream> 7 #include <cmath> 8 #include <cstring> 9 #include <string> 10 11 using namespace std; 12 int n; 13 int sg(int l,int s,int e)//0的个数，s首端点，e末端点 14 { 15 if(s==0 ||e==0)//端点有一边为0一边不为零的返回0的个数 16 { 17 return l; 18 } 19 else 20 return s==e;//端点相同的先手胜返回1，否则后手胜返回0 21 } 22 string g; 23 int main() 24 { 25 cin>>n>>g; 26 bool flag=true; 27 for(int j=0; j<g.size(); j++) 28 if(g[j]!='0') 29 flag=false; 30 if(flag)//全是0的情况，0先手胜,00后手胜,000先手胜，依次循环 31 { 32 if(g.size()%2) 33 { 34 cout<<"FIRST"<<endl; 35 } 36 else 37 cout<<"SECOND"<<endl; 38 return 0; 39 } 40 int b=0; 41 for(int j=0; j<g.size(); j++) 42 { 43 if(g[j]=='0') 44 { 45 int st=j; 46 int s,e; 47 if(j==0) 48 s=0; 49 else 50 s=g[j-1]-'0'; 51 for(; j<g.size(); j++) 52 { 53 if(j+1==g.size()) 54 { 55 e=0; 56 b^=sg(j-st+1,s,e); 57 // cout<<b<<endl; 58 break; 59 } 60 else if(g[j+1]!='0') 61 { 62 e=g[j+1]-'0'; 63 b^=sg(j-st+1,s,e); 64 // cout<<b<<endl; 65 break; 66 } 67 } 68 } 69 } 70 if(b!=0) 71 cout<<"FIRST"<<endl; 72 else 73 cout<<"SECOND"<<endl; 74 } View Code sample input sample output 5 00100 SECOND sample input sample output 4 1020
FIRST
转载于:https://www.cnblogs.com/Skyxj/p/3205351.html
