题集 (搜索扩展)

A 胜利大逃亡(续)（HDU-1429）

（广搜+状态压缩）
题目链接

该题状态压缩拿到的钥匙的情况，将拿到’a’ -'j’的状态分别压缩到1-10位上，vis[i][j][k]用来表示第i行第j列拿到的钥匙状态为k时是否走过，然后进行广搜就可以了.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int M = 25;
const int dir[4][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };

int n, m, t, startx, starty, endx, endy;
char map1[M][M];
// vis[i][j][k] 表示第i行第j列拿到的钥匙状态为k时是否走过
bool vis[M][M][1 << 12];
struct node {
	int x, y, key, time;
}now, next1;

bool check(node nn) {
	if (nn.x < 0 || nn.x >= n || nn.y < 0 || nn.y >= m || map1[nn.x][nn.y] == '*' || vis[nn.x][nn.y][nn.key]) return false;
	return true;
}
// 广搜
void bfs() {
	now.x = startx;
	now.y = starty;
	now.time = 0;
	now.key = 0;
	queue<node> que;
	que.push(now);
	vis[now.x][now.y][now.key] = 1;
	while (!que.empty()) {
		now = que.front();
		que.pop();
		if (now.x == endx && now.y == endy) return;
		for (int i = 0; i < 4; ++i) {
			next1.x = now.x + dir[i][0];
			next1.y = now.y + dir[i][1];
			next1.time = now.time + 1;
			next1.key = now.key;
			if (next1.time >= t) break;
			if (check(next1)) {
				if (map1[next1.x][next1.y] == '.') {
					vis[next1.x][next1.y][next1.key] = 1;
					que.push(next1);
				}
				else if (map1[next1.x][next1.y] >= 'a' && map1[next1.x][next1.y] <= 'j') {
					// 拿到钥匙时将当前状态加上这一步拿到的钥匙
					next1.key = next1.key | (1 << (map1[next1.x][next1.y] - 'a'));
					vis[next1.x][next1.y][next1.key] = 1;
					que.push(next1);
				}
				else if (map1[next1.x][next1.y] >= 'A' && map1[next1.x][next1.y] <= 'Z') {
					// 如果是到达门情况需要检查是否拿到相应的钥匙
					if (next1.key & (1 << (map1[next1.x][next1.y] - 'A'))) {
						vis[next1.x][next1.y][next1.key] = 1;
						que.push(next1);
					}
				}
				else if (map1[next1.x][next1.y] == '@') {
					// 有可能返回起点
					vis[next1.x][next1.y][next1.key] = 1;
					que.push(next1);
				}
				else if (map1[next1.x][next1.y] == '^') {
					vis[next1.x][next1.y][next1.key] = 1;
					que.push(next1);
				}
			}
		}
	}
}

int main() {
	while (scanf("%d%d%d", &n, &m, &t) != EOF) {
		for (int i = 0; i < n; ++i) {
			getchar();
			for (int j = 0; j < m; ++j) {
				map1[i][j] = getchar();
				if (map1[i][j] == '@') startx = i, starty = j;
				else if (map1[i][j] == '^') endx = i, endy = j;
			}
		}
		bfs();
		if (now.x == endx && now.y == endy) printf("%d\n", now.time);
		else printf("-1\n");
		memset(vis, 0, sizeof(vis));
	}
	return 0;
}

B Don’t Get Rooked (POJ-1315)

(DFS暴力递归)
题目链接

该题最大4*4直接暴力递归就行

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int M = 5;

int n;
char map1[M][M];
int vis[M][M];

void draw1(int x, int y) {
	vis[x][y]++;
	for (int i = x + 1; i < n; ++i) {
		if (map1[i][y] == 'X') break;
		vis[i][y] ++;
	}
	for (int i = x - 1; i >= 0; --i) {
		if (map1[i][y] == 'X') break;
		vis[i][y]++;
	}
	for (int i = y + 1; i < n; ++i) {
		if (map1[x][i] == 'X') break;
		vis[x][i]++;
	}
	for (int i = y - 1; i >= 0; --i) {
		if (map1[x][i] == 'X') break;
		vis[x][i]++;
	}
}


void draw0(int x, int y) {
	vis[x][y]--;
	for (int i = x + 1; i < n; ++i) {
		if (map1[i][y] == 'X') break;
		vis[i][y]--;
	}
	for (int i = x - 1; i >= 0; --i) {
		if (map1[i][y] == 'X') break;
		vis[i][y]--;
	}
	for (int i = y + 1; i < n; ++i) {
		if (map1[x][i] == 'X') break;
		vis[x][i]--;
	}
	for (int i = y - 1; i >= 0; --i) {
		if (map1[x][i] == 'X') break;
		vis[x][i]--;
	}
}

int dfs(int x, int y, int num) {
	int ans = num;
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			if (map1[i][j] == 'X') continue;
			if (vis[i][j]) continue;
			draw1(i, j);
			ans = max(ans, dfs(i, j, num + 1));
			draw0(i, j);
		}
	}
	return ans;
}

int main() {
	while (scanf("%d", &n) != EOF) {
		if (n == 0) break;
		for (int i = 0; i < n; ++i) {
			getchar();
			for (int j = 0; j < n; ++j) {
				map1[i][j] = getchar();
			}
		}
		int ans = 0;
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < n; ++j) {
				if (map1[i][j] == 'X') continue;
				draw1(i, j);
				ans = max(ans, dfs(i, j, 1));
				draw0(i, j);
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}

C Igor In the Museum（CodeForces - 598D）

DFS+联通块
题目链接

该题是DFS加联通块,将’.'联通起来，每次dfs一个联通块，最后把需要查询的点对应的块输出出来就行了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int M = 1010;
const int dir[4][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };

int n, m, k;
char map1[M][M];
int ans[M][M];
int num[M * M];
bool vis[M][M];

bool check(int x, int y) {
	if (x <= 0 || x > n || y <= 0 || y > m) return false;
	return true;
}

int dfs(int x, int y, int cnt, int step) {
	int nx, ny;
	ans[x][y] = cnt;
	for (int i = 0; i < 4; ++i) {
		nx = x + dir[i][0];
		ny = y + dir[i][1];
		if (!check(nx, ny)) continue;
		if (map1[nx][ny] == '*') {
			step++;
			continue;
		}
		if (vis[nx][ny]) continue;
		vis[nx][ny] = 1;
		step = dfs(nx, ny, cnt, step);
	}
	return step;
}

int main() {
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 1; i <= n; ++i) {
		getchar();
		for (int j = 1; j <= m; ++j) {
			map1[i][j] = getchar();
		}
	}
	int cnt = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			if (vis[i][j] || map1[i][j] == '*') continue;
			vis[i][j] = 1;
			num[cnt++] = dfs(i, j, cnt, 0);
		}
	}
	int a, b;
	for (int i = 0; i < k; ++i) {
		scanf("%d%d", &a, &b);
		printf("%d\n", num[ans[a][b]]);
	}
	return 0;
}