제출 #97946

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
97946		SpeedOfMagic	경찰관과 강도 (BOI14_coprobber)	C++17	60 / 100	1583 ms	226952 KiB

coprobber

#include "coprobber.h"
#include <bits/stdc++.h>
using namespace std;
 
const int maxN = 500;
const int M = 2 * maxN * maxN;
vector<int> g[maxN];
vector<int> g2[M], rev2[M];
 
//0 if it's officer's turn, 1 if robber's one
int enc(int officerPos, int robberPos, int whoseTurn) {
	return whoseTurn * maxN * maxN + officerPos * maxN + robberPos;
}
 
char vis[M];
int good[M];
int d[M];

void expand(int cur) {
	if (cur / (maxN * maxN) && d[cur] == g2[cur].size()) 
		good[cur] = g2[cur].empty() ? -2 : g2[cur][0];
	else if (cur / (maxN * maxN) == 0 && (d[cur] || g2[cur].size() == 0)) {
		good[cur] = -2;
		for (int i : g2[cur])
			if (good[i] != -1) {
				good[cur] = i;
				break;
			}
	}
	
	if (good[cur] != -1)
		for (int i : rev2[cur])
			if (good[i] == -1) {
				d[i]++;
				expand(i);
			}
}

void add(int v, int u) {
	g2[v].push_back(u);
	rev2[u].push_back(v);
}
 
int loc;
int start(int N, bool A[MAX_N][MAX_N]) {
	memset(good, -1, sizeof good);
	memset(d, 0, sizeof d);
	for (int i = 0; i < N; i++)
		for (int j = 0; j < N; j++)
			if (A[i][j]) 
				g[i].push_back(j);
			
	for (int i = 0; i < N; i++)
		for (int j = 0; j < N; j++) 
			if (i != j) {
				add(enc(i, j, 0), enc(i, j, 1));
				for (int k : g[i])
					add(enc(i, j, 0), enc(k, j, 1));
				for (int k : g[j])
					add(enc(i, j, 1), enc(i, k, 0));
			}
	
	for (int i = 0; i < N; i++) {
		expand(enc(i, i, 0));
		expand(enc(i, i, 1));
	}		
			
	for (int i = 0; i < N; i++) {
		bool no = 0;
		for (int j = 0; j < N; j++)
			no |= good[enc(i, j, 0)] == -1;
		
		if (!no) {
			loc = i;
			return i;
		}
	}
    return -1;
}
 
int nextMove(int R) {
	//cout << loc << " " << R << endl;
	loc = good[enc(loc, R, 0)] % (maxN * maxN) / maxN;
	return loc;
}

컴파일 시 표준 에러 (stderr) 메시지

coprobber.cpp: In function 'void expand(int)':
coprobber.cpp:20:36: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
  if (cur / (maxN * maxN) && d[cur] == g2[cur].size()) 
                             ~~~~~~~^~~~~~~~~~~~~~~~~

• Subtask 1: 16 / 16
• Subtask 2: 14 / 14
• Subtask 3: 30 / 30
• Subtask 4: 0 / 40