hdu 4678 Mine（bfs+博弈）

解题思路

代码

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>

using namespace std;
typedef pair<int,int> pii;
const int maxn = 1005;
const int dir[][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};

int N, M, K, G[maxn][maxn], T[maxn][maxn];

void init () {
scanf("%d%d%d", &N, &M, &K);
memset(G, 0, sizeof(G));
memset(T, 0, sizeof(T));

int u, v;
for (int i = 0; i < K; i++) {
scanf("%d%d", &u, &v);
G[u][v] = -1;
for (int j = 0; j < 8; j++) {
int x = u + dir[j][0];
int y = v + dir[j][1];
if (x < 0 || x >= N || y < 0 || y >= M || G[x][y] == -1) continue;
G[x][y] = 1;
}
}
}

queue<pii> Q;

int bfs(int x, int y) {
int ret = 0;
Q.push(make_pair(x, y));
T[x][y] = 1;

while (!Q.empty()) {
x = Q.front().first;
y = Q.front().second;
Q.pop();

if (G[x][y]) { ret++; continue; }

for (int i = 0; i < 8; i++) {
int p = x + dir[i][0];
int q = y + dir[i][1];
if (p < 0 || p >= N || q < 0 || q >= M || T[p][q]) continue;
Q.push(make_pair(p, q));
T[p][q] = 1;
}
}
return ret&1 ? 2 : 1;
}

int main () {
int cas;
scanf("%d", &cas);
for (int kcas = 1; kcas <= cas; kcas++) {
init();

int ans = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (G[i][j] == 0 && T[i][j] == 0)
ans ^= bfs(i, j);
}
}
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
if (G[i][j] == 1 && T[i][j] == 0)
ans ^= 1;
}
printf("Case #%d: %s\n", kcas, ans ? "Xiemao" : "Fanglaoshi");
}
return 0;
}