제출 #24462

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
24462		noobprogrammer	경찰관과 강도 (BOI14_coprobber)	C++14	100 / 100	1069 ms	11256 KiB

coprobber

#include<bits/stdc++.h>
#include "coprobber.h"
using namespace std;
typedef long long ll;
#define fi first
#define se second
#define ii pair<int,int>
#define vii vector<pair<int,int> >
#define vi vector<int>
// const int MAX_N = 500 ;

struct node{
    int  co , ro , nw ;
} ;

bool wn[505][505][2] ;
int lft[505][505][2] , nmv[505][505][2] , visited[505][505][2] ; // [cop][rob] ; 1 = rob ; 0 =  cop
int pos = -1 ;

int start(int n, bool A[MAX_N][MAX_N])
{
    queue<node> q ;
    for(int i=0;i<n;i++){
        int cnt = 0 ;
        for(int j=0;j<n;j++) cnt+= (int)A[i][j] ;
        for(int j=0;j<n;j++){
            lft[i][j][0] = 1 ;
            lft[j][i][1] = cnt ;
        }
        for(int j=0;j<2;j++ ) wn[i][i][j] = 1 , q.push({i,i,j}) , lft[i][i][j] = 0 , nmv[i][i][j] = i ;
    }
    int r , c , tr , nxt ;
    while(!q.empty()){
        r = q.front().ro ; c = q.front().co ; tr = q.front().nw ; nxt = 1-tr ;
        q.pop() ;
        if(visited[c][r][tr]) continue ;
        visited[c][r][tr] = 1 ;
        if(!tr){
            for(int i = 0 ;i<n;i++){
                if(!A[r][i]) continue ;
                if(visited[c][i][nxt]) continue ;
                lft[c][i][nxt]-- ;
                if(lft[c][i][nxt]) continue ;
                wn[c][i][nxt] = 1  ;
                q.push({c,i,nxt}) ;
            }
        }
        else{
            for(int i = 0 ;i<n;i++){
                if(!A[c][i] && i!=c ) continue ;
                if(visited[i][r][nxt]) continue ;
                lft[i][r][nxt]-- ;
                if(lft[i][r][nxt]) continue ;
                wn[i][r][nxt] = 1 ;
                nmv[i][r][nxt] = c ;
                q.push({i,r,nxt}) ;
            }
        }
    }

    for(int i=0;i<n;i++) {
        bool chk = 1 ;
        for(int j=0;j<n;j++) chk = chk & wn[i][j][1] ;
        if(!chk) continue ;
        pos = i ;
        break ;
    }
    return pos ;
}

int nextMove(int R)
{
    return pos = nmv[pos][R][0] ;
}

• Subtask 1: 16 / 16
• Subtask 2: 14 / 14
• Subtask 3: 30 / 30
• Subtask 4: 40 / 40