Code Forces 786A Berzerk(有向图博弈)

  1 #include <iostream>
  2 #include <queue>
  3 #include <stack>
  4 #include <cstdio>
  5 #include <vector>
  6 #include <map>
  7 #include <set>
  8 #include <bitset>
  9 #include <algorithm>
 10 #include <cmath>
 11 #include <cstring>
 12 #include <cstdlib>
 13 #include <string>
 14 #include <sstream>
 15 #include <time.h>
 16 #define x first
 17 #define y second
 18 #define pb push_back
 19 #define mp make_pair
 20 #define lson l,m,rt*2
 21 #define rson m+1,r,rt*2+1
 22 #define mt(A,B) memset(A,B,sizeof(A))
 23 using namespace std;
 24 typedef long long LL;
 25 //const double PI = acos(-1);
 26 const int N=7e3+10;
 27 const LL mod=1e9+7;
 28 const int inf = 0x3f3f3f3f;
 29 const LL INF=0x3f3f3f3f3f3f3f3fLL;
 30 int step[N][2];
 31 int vis[N][2];
 32 int du[N][2];
 33 int k[2];
 34 int n;
 35 struct node
 36 {
 37     int x,y;
 38     int type;
 39 };
 40 queue<struct node> Q;
 41 void bfs()
 42 {
 43     struct node p,q;
 44     while(!Q.empty())
 45     {
 46         p=Q.front();
 47         Q.pop();
 48         if(!p.type)
 49         {
 50             int turn=!p.y;
 51             for(int i=0;i<k[turn];i++)
 52             {
 53                 int bu=(p.x+n-step[i][turn])%n;
 54                 //cout<<p.x<<" "<<bu<<endl;
 55                 if(vis[bu][turn]==-1)
 56                 {
 57                     vis[bu][turn]=1;
 58                     Q.push({bu,turn,1});
 59                 }
 60             }
 61         }
 62         else
 63         {
 64             int turn=!p.y;
 65             for(int i=0;i<k[turn];i++)
 66             {
 67                 int bu=(p.x+n-step[i][turn])%n;
 68                 du[bu][turn]--;
 69                 if(du[bu][turn]<=0&&vis[bu][turn]==-1)
 70                 {
 71                     vis[bu][turn]=0;
 72                     Q.push({bu,turn,0});
 73                 }
 74             }
 75         }
 76     }
 77 }
 78 int main()
 79 {
 80 #ifdef Local
 81     freopen("data.txt","r",stdin);
 82 #endif
 83     ios::sync_with_stdio(false);
 84     cin.tie(0);
 85     mt(vis,-1);
 86     cin>>n;
 87     cin>>k[0];
 88     for(int i=0;i<k[0];i++)cin>>step[i][0];
 89     for(int i=0;i<n;i++)du[i][0]=k[0];
 90     cin>>k[1];
 91     for(int i=0;i<k[1];i++)cin>>step[i][1];
 92     for(int i=0;i<n;i++)du[i][1]=k[1];
 93     vis[0][1]=vis[0][1]=0;//失败
 94     Q.push({0,0,0});
 95     Q.push({0,1,0});
 96     bfs();
 97     for(int d=0;d<2;d++)
 98     {
 99         for(int i=1;i<n;i++)
100         {
101             if(vis[i][d]==0)cout<<"Lose ";
102             else if(vis[i][d]==1)cout<<"Win ";
103             else cout<<"Loop ";
104         }
105         cout<<endl;
106     }
107 
108 #ifdef Local
109     cerr << "time: " << (LL) clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
110 #endif
111 }
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原文地址:https://www.cnblogs.com/27sx/p/6653114.html