| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 90427 | nikolapesic2802 | 경찰관과 강도 (BOI14_coprobber) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*
-Make a table value[i][j][move], i meaning where the cop is located, j meaning where the robber is located and move telling us who needs to make the next move.
-A state is winning if:
*i==j
*Its the cops move and he can move to a winning state
*Its the robbers move and he cant avoid going to a winning state
-Calculate the values that we know for sure in this table(there might me a cycle in the graph so we cant tell if a position is winning or losing, in that case we can just think of it as a losing position)
-For every cops move calculate the move that leads us to a winning state
-All states need to be winning in order to be able to catch the cop.
-Also if we can catch a cop starting from one location, we can do it from any location.
*/
#include <bits/stdc++.h>
#include "coprobber.h"
using namespace std;
#define pb push_back
vector<vector<int> > graph(MAX_N);
int value[MAX_N][MAX_N][2],nxt[MAX_N][MAX_N],degree[MAX_N][MAX_N],pos=0;
int start(int N, bool A[MAX_N][MAX_N])
{
for(int i=0;i<N;i++)
for(int j=0;j<N;j++)
if(A[i][j])
graph[i].pb(j),degree[i]++,degree[j]++;
queue<vector<int> > q;
for(int i=0;i<N;i++)
value[i][i][0]=1,value[i][i][1]=1,q.push({i,i,0}),q.push({i,i,1});
while(q.size())
{
vector<int> d=q.front();q.pop();
int x=d[0],y=d[1],z=d[2];
if(z==1)
{
int i=x,j=y,k=0;
if(value[i][j][k]==0)
{
value[i][j][k]=1;
nxt[i][j]=x;
q.push({i,j,k});
}
for(auto p:graph[x])
{
i=p;j=y;k=0;
if(value[i][j][k]==0)
{
value[i][j][k]=1;
nxt[i][j]=x;
q.push({i,j,k});
}
}
}
else
{
for(auto p:graph[y])
{
int i=x,j=p,k=1;
if(value[i][j][k]==0)
{
degree[i][j]--;
if(degree[i][j]==0)
{
value[i][j][k]=1;
q.push({i,j,k});
}
}
}
}
}
bool t=true;
for(int i=1;i<N;i++)
if(value[0][i][0]!=1)
t=false;
if(t)
return 0;
return -1;
}
int nextMove(int R)
{
return pos=nxt[pos][R];
}