【kuangbin专题一】简单搜索

[2020.12.21 ]

自从上个礼拜打铁以来，感觉整个人都不对劲。

最近几天打了一些比赛，找回了一点感觉。

这段时间到考试前，我打算刷一刷kuangbin题单，毕竟这是开学的时候就立下的Flag...

...只完成了最后一个不用脑子的...

实在是太草了

一、 搜索专题

A - 棋盘问题

#include<cstdio>
using namespace std;
const int N = 1e5 + 10;

int n, k, ans;
char mp[10][10];
int col[10];

void dfs(int now,int sum){
	if(sum == k){
		ans++;return;
	}
	if (now > n)return;
	
	for(int i = 1;i <= n;i++){
		if(mp[now][i] == '#' and !col[i]){
			col[i] = 1;
			dfs(now + 1, sum + 1);
			col[i] = 0;
		}
	}
	dfs(now + 1, sum);
}
int main(){
	while(~scanf("%d%d",&n,&k) and n != -1){
		for (int i = 1;i <= n;i++) scanf("%s", mp[i] + 1);
		ans = 0;
		dfs(0, 0);
		printf("%d
", ans);
	}
}

B - Dungeon Master

三维迷宫bfs，我属实有点拉跨了。

我开始一直MLE还埋怨是不是poj的问题，后来发现自己bfs没有开vis数组...

后来输出的时候没有加 . （一个优雅的英文句号） ...

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-21 15:01:56
 */
#include<cstdio>
#include<queue>
using namespace std;

int h, n, m;

char mp[40][40][40];
typedef struct {
    int h, x, y;
}p;

const int dh[] = { 1,-1,0,0,0,0 };
const int dx[] = { 0,0,1,-1,0,0 };
const int dy[] = { 0,0,0,0,1,-1 };

p st, ed;

int main() {
	while (scanf("%d%d%d", &h, &n, &m) and h != 0) {
		for (int i = 1; i <= h; i++) {
			for (int j = 1; j <= n; j++) {
				scanf("%s", mp[i][j] + 1);
				for (int k = 1; k <= m; k++) {
					if (mp[i][j][k] == 'S')st = { i,j,k };
					if (mp[i][j][k] == 'E')ed = { i,j,k };
				}
			}
		}
		queue<p>Q; Q.push(st);
		int ans = -1, now = 0;
		while (not Q.empty()) {
			int sz = Q.size();
			while (sz--) {
				p tp = Q.front(); Q.pop();
				if (tp == ed) {
					ans = now;
					goto END;
				}
				for (int i = 0; i < 6; i++) {
					int nh = tp.h + dh[i];
					int nx = tp.x + dx[i];
					int ny = tp.y + dy[i];
					if (nh > h or nh < 1 or nx > n or nx < 1 or ny > m or ny < 1)continue;
					if (mp[nh][nx][ny] == '#')continue;
					Q.push({ nh,nx,ny });
				}
			}
			now++;
		}
	END:
		if (ans == -1)puts("Trapped!");
		else printf("Escaped in %d minute(s).
", ans);
	}
}

C - Catch That Cow

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-21 15:19:05
 */
#include<cstdio>
#include<queue>
using namespace std;
int n, k;
int vis[100010];
int main() {
	scanf("%d%d", &n, &k);
	queue<int>Q; Q.push(n); vis[n] = 1;
	int ans = 0;
	while (not Q.empty()) {
		int sz = Q.size();
		while (sz--) {
			int tp = Q.front(); Q.pop();
			if (tp == k) {
				printf("%d
", ans);
				return 0;
			}
			if (tp > 0 and !vis[tp - 1])vis[tp - 1] = 1, Q.push(tp - 1);
			if (tp < 100009 and !vis[tp + 1])vis[tp + 1] = 1, Q.push(tp + 1);
			if (tp * 2 < 100010 and !vis[tp * 2])vis[tp * 2] = 1, Q.push(2 * tp);
		}
		ans++;
	}
	puts("-1");

}

​

D - Fliptile

很垃圾的一份代码，只有一个剪枝。

代码也写的没法看...

我不应该瞧不起这些题目...

其实好像第一行确定了方案就确定了，好像可以暴力枚举第一行

好像也差不多

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-21 15:33:28
 * @res: 641ms / limit 2000ms
 */

#include<cstdio>

bool flag;
int mp[20][20], n, m;
const int dx[] = { 1,0,-1,0,0 };
const int dy[] = { 0,1,0,-1,0 };
int ans[20][20], res[20][20], mx;
void dfs(int x, int y,int sum) {
	if (x > n) {
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++)if (mp[i][j] == 1)return;
		}
		flag = true;
		if (sum < mx) {
			mx = sum;
			for (int i = 1; i <= n; i++) {
				for (int j = 1; j <= m; j++) {
					res[i][j] = ans[i][j];
					//printf("%d%c", ans[i][j], " 
"[j == m]);
				}
			}
		}
		else if (sum == mx) {
			bool le = true;
			for (int i = 1; i <= n; i++) {
				for (int j = 1; j <= m; j++) {
					if (ans[i][j] > res[i][j])le = false;
				}
			}
			if (le) {
				for (int i = 1; i <= n; i++) {
					for (int j = 1; j <= m; j++) {
						res[i][j] = ans[i][j];
						//printf("%d%c", ans[i][j], " 
"[j == m]);
					}
				}
			}
		}
		return;
	}
	if (x > 1) {
		if (mp[x - 1][y] == 0) {
			//if (x == n and y > 1 and mp[x][y - 1] == 1)return;
			ans[x][y] = 0, dfs(x + (y == m), (y == m) ? 1 : y + 1, sum);
		}
		else {
			//if (x == n && y > 1 && mp[x][y - 1] == 0)return;
			ans[x][y] = 1;
			for (int i = 0; i < 5; i++)mp[x + dx[i]][y + dy[i]] ^= 1;
			dfs(x + (y == m), (y == m) ? 1 : y + 1, sum + 1);
			for (int i = 0; i < 5; i++)mp[x + dx[i]][y + dy[i]] ^= 1;
		}
	}
	else {
		ans[x][y] = 0;
		dfs(x + (y == m), (y == m) ? 1 : y + 1, sum);

		ans[x][y] = 1;
		for (int i = 0; i < 5; i++)mp[x + dx[i]][y + dy[i]] ^= 1;
		dfs(x + (y == m), (y == m) ? 1 : y + 1, sum + 1);
		for (int i = 0; i < 5; i++)mp[x + dx[i]][y + dy[i]] ^= 1;
	}
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)scanf("%1d", mp[i] + j);
	mx = 1000000;
	dfs(1, 1, 0);
	if (!flag)puts("IMPOSSIBLE");
	else {
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				//res[i][j] = ans[i][j];
				printf("%d%c", res[i][j], " 
"[j == m]);
			}
		}
	}
}

我一番压行操作后

我悟出一个道理

在DFS的时候，在进入dfs前剪枝和进入dfs后再剪枝是不一样的，前则要快不少。

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-21 15:59:09
 * @res: 782ms / limit: 2000ms
 */

#include<cstdio>
#define max(a,b) (a>b?a:b)

bool flag;
int mp[20][20], n, m;
const int dx[] = { 1,0,-1,0,0 };
const int dy[] = { 0,1,0,-1,0 };
int ans[20][20], res[20][20], mx;
void dfs(int x, int y, int sum) {
	if (x > n) {
		if (sum > mx)return;
		for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)if (mp[i][j] == 1 || (sum == mx && (ans[i][j] > res[i][j])))return;
		flag = true; mx = sum;
		for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)res[i][j] = ans[i][j];
		return;
	}
	if(x == 1 || mp[x-1][y] == 0)ans[x][y] = 0,dfs(x + (y == m), (y == m) ? 1 : y + 1, sum);
	if (x == 1 || mp[x - 1][y] == 1) {
		for (int i = 0; i < 5; i++)mp[x + dx[i]][y + dy[i]] ^= 1;
		ans[x][y] = 1;
		dfs(x + (y == m), (y == m) ? 1 : y + 1, sum + 1);
		for (int i = 0; i < 5; i++)mp[x + dx[i]][y + dy[i]] ^= 1;
	}
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)scanf("%1d", mp[i] + j);
	mx = 751404976; dfs(1, 1, 0);
	if (!flag)puts("IMPOSSIBLE");
	else {
		for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)printf("%d%c", res[i][j], " 
"[j == m]);
	}
}

E - Find The Multiple

暴力算了一些数据发现位数都比较小，于是直接从低位搜到高位。

裸的搜索可以直接交上去，跑了 938ms

发现 m == 198 的时候搜的答案是 1111111111111111110，可以直接加个特判。

跑了 344ms

当然也可以直接打表光速过题。

#include<cstdio>
/* status : 938ms / 1000ms */
using namespace std;

typedef long long ll;
ll ans[300];
ll get(ll j) {
    if (ans[j])return ans[j];
    for (int i = 1;i < 1 << 20;i++) {
        ll base = 1, now = 0;
        for (int k = 0;k < 20;k++) {
            if ((i >> k) & 1)now = (now + base) % j;
            base = base * 10 % j;
        }
        if (now == 0)return ans[j] = i;
    }
}
void print(ll x) {
    if (!x)return;
    print(x >> 1);
    if (x & 1)putchar('1');else putchar('0');
}
int main() {
    int n;
    while (scanf("%d", &n) && n > 0) {
        print(get(n));puts("");
    }
}

#include<cstdio>
/* 加了特判  status: 344ms / 1000ms  */
using namespace std;

typedef long long ll;
ll ans[300];
ll get(ll j) {
	if (ans[j])return ans[j];
	for (int i = 1; i < 1 << 19; i++) {
		ll base = 1, now = 0;
		for (int k = 0; k < 19; k++) {
			if ((i >> k) & 1)now = (now + base) % j;
			base = base * 10 % j;
		}
		if (now == 0)return ans[j] = i;
	}
}
void print(ll x) {
	if (!x)return;
	print(x >> 1);
	if (x & 1)putchar('1'); else putchar('0');
}
int main() {
	int n;
	while (scanf("%d", &n) && n > 0) {
		if (n == 198) {
			puts("1111111111111111110"); continue;
		}
		print(get(n)); puts("");
	}
}

F - Prime Path

简单bfs，先建边就可以

#include<cstdio>
#include<queue>
#include<map>
#include<vector>

using namespace std;

bool judge(int x) {
	for (int i = 2; i * i <= x; i++)if (x % i == 0)return false;
	return true;
}
bool ok(int x, int y) {
	int dif = 0;
	for (int i = 0; i < 4; i++) {
		if (x % 10 != y % 10)dif++;
		x /= 10; y /= 10;
	}
	return dif == 1;
}
int T, x, y;
map<int, vector<int> >G;
int main() {
	vector<int>L;
	for (int i = 1000; i <= 9999; i++) {
		if (judge(i))L.push_back(i);
	}
	int n = L.size();
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < i; j++) {
			if (ok(L[i], L[j])) {
				G[L[i]].push_back(L[j]);
				G[L[j]].push_back(L[i]);
			}
		}
	}
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d", &x, &y);
		queue<int>Q;
		int ans = -1, now = 0;
		Q.push(x); map<int, int>vis; vis[x] = 1;

		while (!Q.empty()) {
			int sz = Q.size();
			while (sz--) {
				int tp = Q.front(); Q.pop();
				if (tp == y) {
					ans = now;
					break;
				}
				for (int i = 0; i < G[tp].size();i++) {
					int v = G[tp][i];
					if (!vis[v]) {
						Q.push(v);
						vis[v] = 1;
					}
				}
			}
			if (ans != -1)break;
			now++;
		}
		if (ans == -1)puts("Impossible");
		else {
			printf("%d
", ans);
		}
	}
}

G - Shuffle'm Up

直接模拟就可以，注意要手动加一个 ''

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-22 09:57:16
 */
#include<cstdio>
#include<cstring>
int T, n;
char s[200], p[200], tm[300], st[300], ed[300];

int main() {
	scanf("%d", &T); int c = 0;
	while (T--) {
		scanf("%d", &n);
		scanf("%s", s + 1);
		scanf("%s", p + 1);
		scanf("%s", ed + 1);
		for (int i = 1; i <= n; i++) {
			st[2 * i - 1] = p[i];
			st[2 * i] = s[i];
		}
		st[2 * n + 1] = 0; //!!!
		strcpy(tm + 1, st + 1);
		int ans = -1, now = 1;
		while (1) {
			for (int i = 1; i <= n; i++) {
				s[i] = tm[i]; p[i] = tm[i + n];
			}
			if (!strcmp(tm + 1, ed + 1)) {
				ans = now;
				break;
			}
			if (!strcmp(tm + 1, st + 1) && now != 1) {
				break;
			}
			for (int i = 1; i <= n; i++) {
				tm[2 * i - 1] = p[i];
				tm[2 * i] = s[i];
			}
			now++;
		}
		printf("%d %d
", ++c, ans);
	}
}

H - Pots

刚开始以为会很难写，其实细细想一想实现起来也不是特别难。

比较锻炼代码能力

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-22 19:27:20
 */
#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
#define min(a,b) (a<b?a:b)
using namespace std;
int A, B, C;
struct st {
	int a, b;
	st() {}
	st(int _a, int _b) { a = _a, b = _b; }
	bool operator < (const st& x)const {
		if (a != x.a)return a < x.a;
		return b < x.b;
	}
}ed;
int vis[200][200];
map<st, st>pre;
char pre_op[200][200][200];
const char op[6][200] = {
	"FILL(1)","FILL(2)","DROP(1)",
	"DROP(2)","POUR(1,2)","POUR(2,1)"
};
st res[6];
void operate(st x) {
	res[0] = st(A, x.b); res[1] = st(x.a, B);
	res[2] = st(0, x.b); res[3] = st(x.a, 0);
	int mx = min(x.a, B - x.b); res[4] = st(x.a - mx, x.b + mx);
	mx = min(A - x.a, x.b); res[5] = st(x.a + mx, x.b - mx);
}
void print(st x) {
	if (x.a == 0 and x.b == 0)return;
	print(pre[x]);
	printf("%s
", pre_op[x.a][x.b]);
}
int main() {
	scanf("%d%d%d", &A, &B, &C);
	queue<st>Q;
	Q.push({ 0,0 }); vis[0][0] = 1;
	int ans = -1, now = 0;
	while (not Q.empty()) {
		int sz = Q.size();
		while (sz--) {
			st tp = Q.front(); Q.pop();
			//printf("(%d , %d)
", tp.a, tp.b);
			if (tp.a == C or tp.b == C) {
				ans = now;
				ed = tp;
				break;
			}
			operate(tp);
			for (int i = 0; i < 6; i++) {
				if (vis[res[i].a][res[i].b])continue;
				vis[res[i].a][res[i].b] = 1;
				pre[res[i]] = tp;
				strcpy(pre_op[res[i].a][res[i].b], op[i]);
				Q.push(res[i]);
			}
		}
		if (ans != -1)break;
		now++;
	}
	if (ans == -1)puts("impossible");
	else {
		printf("%d
", ans);
		print(ed);
	}
}

I - Fire Game

看到数据量这么小，脑子里不要带东西。

直接用最裸的暴力就可以了，不需要任何优化

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-22 20:00:09
 */
#include<cstdio>
#include<queue>
#include<cstring>
#define min(a,b) (a<b?a:b)
using namespace std;
int T, n, m;
char mp[20][20];
struct point {
	int x, y;
	point() {}
	point(int _x, int _y) {
		x = _x, y = _y;
	}
};
int vis[20][20];
const int dx[] = { 1,0,-1,0 };
const int dy[] = { 0,1,0,-1 };

int cal(point st1, point st2) {
	int ans = 0; memset(vis, 0, sizeof vis);
	queue<point>Q; Q.push(st1); Q.push(st2);
	vis[st1.x][st1.y] = vis[st2.x][st2.y] = 1;
	while (!Q.empty()) {
		int sz = Q.size();
		while (sz--) {
			point tp = Q.front(); Q.pop();
			for (int i = 0; i < 4; i++) {
				int nx = tp.x + dx[i], ny = tp.y + dy[i];
				if (nx < 1 or nx > n or ny < 1 or ny > m or mp[nx][ny] != '#' or vis[nx][ny])continue;
				vis[nx][ny] = 1;
				Q.push(point(nx, ny));
			}
			
		}ans++;
	}
	for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)if (mp[i][j] == '#' and !vis[i][j])return 0x3f3f3f3f;
	return ans - 1;
}
int main() {
	scanf("%d", &T); int c = 0;
	while (T--) {
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++)scanf("%s", mp[i] + 1);
		int ans = 0x3f3f3f3f;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				if (mp[i][j] != '#')continue;
				for (int x = 1; x <= n; x++) {
					for (int y = 1; y <= m; y++) {
						if (mp[x][y] != '#')continue;
						ans = min(ans, cal(point(i, j), point(x, y)));
					}
				}
			}
		}
		if (ans == 0x3f3f3f3f)ans = -1;
		printf("Case %d: %d
", ++c, ans);
	}
}

J - Fire!

先bfs出每个点到火源的距离，再对joe进行bfs一次，若距离joe的距离大于等于到火的距离则不能入队，看能否到达边缘就可以，Uva挂了不知道对没对

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-23 10:41:10
 */
#include<bits/stdc++.h>
using namespace std;

int T, n, m;
char mp[1010][1010];

struct p {
	int x, y;
};
p fire, joe;
//map<p, int>dis, vis;
int dis[1010][1010], vis[1010][1010];
const int dx[] = { 1,0,-1,0 };
const int dy[] = { 0,1,0,-1 };

int main() {
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d", &n, &m);
		queue<p>Q; 
		memset(dis, 0, sizeof dis);
		memset(vis, 0, sizeof vis);
		for (int i = 1; i <= n; i++) {
			scanf("%s", mp[i] + 1);
			for (int j = 1; j <= m; j++) {
				if (mp[i][j] == 'J')joe = { i,j };
				if (mp[i][j] == 'F')Q.push({ i,j }), vis[i][j] = 1;
			}
		}
		
		while (not Q.empty()) {
			int sz = Q.size();
			while (sz--) {
				p tp = Q.front(); Q.pop();
				for (int i = 0; i < 4; i++) {
					int nx = tp.x + dx[i], ny = tp.y + dy[i];
					if (nx < 1 or nx > n or ny < 1 or ny > m or mp[nx][ny] == '#' or vis[nx][ny])continue;
					vis[nx][ny] = 1;
					dis[nx][ny] = dis[tp.x][tp.y] + 1;
					Q.push({ nx,ny });
				}
			}
		}
		int ans = -1, now = 0; 
		Q.push(joe); memset(vis,0,sizeof vis); vis[joe.x][joe.y] = 1;
		while (not Q.empty()) {
			int sz = Q.size();
			while (sz--) {
				p tp = Q.front(); Q.pop();
				if (tp.x == 1 or tp.x == n or tp.y == 1 or tp.y == m) {
					ans = now;
					break;
				}
				for (int i = 0; i < 4; i++) {
					int nx = tp.x + dx[i], ny = tp.y + dy[i];
					if (mp[nx][ny] == '#' or (dis[nx][ny] != 0 and now + 1 >= dis[nx][ny]) or vis[nx][ny])continue;
					vis[nx][ny] = 1;
					Q.push({ nx,ny });
				}
			}
			if (ans != -1)break;
			now++;
		}
		if (ans == -1)puts("IMPOSSIBLE");
		else printf("%d
", ans + 1);
	}
}

K - 迷宫问题

这个题我开始觉得太水了，不想写，结果debug，滴了半个多小时

且听我细细道来

对于样例

0 1 0 0 0
0 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0

我在下面的代码片加了断点，发现当 nxt == {3,2} 的时候，被 continue 略过去了

我开始改成了 vis.count(nxt) 发现没啥变化，我把 vis[nxt] 去掉，就没有被略过去

int nx = tp.x + dx[i], ny = tp.y + dy[i];
p nxt = { nx,ny };
if (nx < 1 or nx > 5 or ny < 1 or ny > 5 or mz[nx][ny] == 1 or vis[nxt])continue;

后来我一想，原来是重载的问题

bool operator < (const point& b)const {
	//return x < b.x;
	return x == b.x ? y < b.y : x < b.x;
}

注释掉的是最开始的，因为最开始的时候map报错了，我意识到是要重载运算符，我就随便重载了一下。

队友说是要满足自反性

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-22 10:51:10
 */
#include<cstdio>
#include<map>
#include<queue>

using namespace std;
int mz[10][10];
typedef struct point {
	int x, y;
	bool operator == (point b) {
		return x == b.x and y == b.y;
	}
	bool operator < (const point& b)const {
		//return x < b.x;
		return x == b.x ? y < b.y : x < b.x;
	}
}p;
p st, ed;
map<p, p>pre;
map<p, int>vis;
const int dx[] = { 1,0,-1,0 };
const int dy[] = { 0,1,0,-1 };
void print(p x) {
	if (x == st) {
		printf("(%d, %d)
", x.x - 1, x.y - 1);
		return;
	}
	print(pre[x]);
	printf("(%d, %d)
", x.x - 1, x.y - 1);
}
int main() {
	for (int i = 1; i <= 5; i++)for (int j = 1; j <= 5; j++)scanf("%d", mz[i] + j);
	st = { 1,1 }, ed = { 5,5 };
	queue<p>Q; Q.push(st); vis[st] = 1;

	while (not Q.empty()) {
		p tp = Q.front(); Q.pop();
		
		if (tp == ed) {
			break;
		}
		for (int i = 0; i < 4; i++) {
			int nx = tp.x + dx[i], ny = tp.y + dy[i];
			p nxt = { nx,ny };

			if (nx < 1 or nx > 5 or ny < 1 or ny > 5 or mz[nx][ny] == 1 or vis.count(nxt) == 1)continue;
			vis[nxt] = 1;
			pre[nxt] = tp;
			Q.push(nxt);
			//printf("%d %d
", nxt.x, nxt.y);
		}
	}
	print(ed);
}

L - Oil Deposits

用个并查集暴力就可以

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-22 17:57:53
 */
#include<cstdio>
#include<set>
#define and &&
#define or ||
using namespace std;
int n, m;

char mp[200][200];
int fa[20000];
int find(int a) {
	return a == fa[a] ? a : fa[a] = find(fa[a]);
}
void merge(int x, int y) {
	x = find(x), y = find(y);
	if (x != y)fa[x] = y;
}
int getid(int x, int y) {
	return (x - 1) * m + y;
}
const int dx[] = { 1,1,0,-1,-1,-1,0,1 };
const int dy[] = { 0,1,1,1,0,-1,-1,-1 };
int main() {
	while (scanf("%d%d", &n, &m) and n) {
		for (int i = 1; i <= n; i++) {
			scanf("%s", mp[i] + 1);
		}

		for (int i = 1; i <= n * m; i++)fa[i] = i;

		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				for (int k = 0; k < 8; k++) {
					int nx = i + dx[k], ny = j + dy[k];
					if (mp[i][j] != '@' or nx < 1 or nx > n or ny < 1 or ny > m or mp[nx][ny] != '@')continue;
					merge(getid(i, j), getid(nx, ny));
				}
			}
		}
		set<int>s;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				if (mp[i][j] == '@')s.insert(find(getid(i, j)));
			}
		}
		printf("%d
", s.size());
	}
}

M - 非常可乐

直接暴力

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-22 18:18:19
 */
#include<cstdio>
#include<queue>
#include<map>
#define min(a,b) (a<b?a:b)
#define not !
#define and &&
#define or ||
using namespace std;

int S, N, M;
typedef struct p {
	int a, b, c;
	p() {}
	p(int _a, int _b, int _c) {
		a = _a, b = _b, c = _c;
	}
	bool operator < (const p& rhs)const {
		if (a != rhs.a)return a < rhs.a;
		if (b != rhs.b)return b < rhs.b;
		return c < rhs.c;
	}
}tup;
map<tup, int>vis;
bool ok(tup x) {
	return (x.a == 0 and x.b == x.c) or (x.b == 0 and x.a == x.c) or (x.c == 0 and x.a == x.b);
}

tup res[6];
void operate(tup x) {
	int mx = min(N - x.b, x.a);
	res[0] = tup(x.a - mx, x.b + mx, x.c);
	mx = min(M - x.c, x.a);
	res[1] = tup(x.a - mx, x.b, x.c + mx);
	mx = min(S - x.a, x.b);
	res[2] = tup(x.a + mx, x.b - mx, x.c);
	mx = min(M - x.c, x.b);
	res[3] = tup(x.a, x.b - mx, x.c + mx);
	mx = min(S - x.a, x.c);
	res[4] = tup(x.a + mx, x.b, x.c - mx);
	mx = min(N - x.b, x.c);
	res[5] = tup(x.a, x.b + mx, x.c - mx);
}
int main() {
	while (scanf("%d%d%d", &S, &N, &M) and S != 0) {

		queue<tup>Q; Q.push(p(S, 0, 0));
		int ans = -1, now = 0;
		vis.clear();
		while (not Q.empty()) {
			int sz = Q.size();
			while (sz--) {
				tup tp = Q.front(); Q.pop();
				if (ok(tp)) {
					ans = now;
					break;
				}
				operate(tp);
				for (int i = 0; i < 6; i++) {
					if (vis.count(res[i]))continue;
					Q.push(res[i]);
					vis[res[i]] = 1;
				}
			}
			if (ans != -1)break;
			now++;
		}
		if (ans == -1)puts("NO");
		else printf("%d
", ans);
	}
}

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-23 10:29:17
 */
#include<cstdio>
#include<cstring>
#include<queue>
#include<bits/stdc++.h>
using namespace std;
#define min(a,b) (a < b ? a : b)
int n, m;
char mp[220][220];
const int dx[] = { 1,0,-1,0 };
const int dy[] = { 0,1,0,-1 };
struct point {
	int x, y;
	point() {}
	point(int _x, int _y) { x = _x, y = _y; }
	bool operator < (const point& b)const {
		if (x != b.x)return x < b.x;
		return y < b.y;
	}
}stA,stB;
int id(point x) {return (x.x - 1) * m + x.y;}
int disA[50000], disB[50000], vis[50000];

void bfs(point st,int* dis) {
	memset(vis, 0, sizeof vis); vis[id(st)] = 1;
	queue<point>Q;Q.push(st);
	int now = 0;
	while (not Q.empty()) {
		int sz = Q.size();
		while (sz--) {
			point tp = Q.front(); Q.pop();
			dis[id(tp)] = now;
			for (int i = 0; i < 4; i++) {
				point nxt = point(tp.x + dx[i], tp.y + dy[i]);
				if (nxt.x < 1 or nxt.x > n or nxt.y < 1 or nxt.y > m or vis[id(nxt)])continue;
				if (mp[nxt.x][nxt.y] == '#')continue;
				vis[id(nxt)] = 1;
				Q.push(nxt);
			}
		}
		now++;
	}
}
int main() {
	while (~scanf("%d%d", &n, &m)) {
		for (int i = 1; i <= n; i++) {
			scanf("%s", mp[i] + 1);
			for (int j = 1; j <= m; j++) {
				if (mp[i][j] == 'M')stA = point(i, j);
				if (mp[i][j] == 'Y')stB = point(i, j);
			}
		}
		memset(disA, 0x3f, sizeof disA); memset(disB, 0x3f, sizeof disB);
		bfs(stA, disA); bfs(stB, disB);
		int ans = 0x3f3f3f3f;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				if (mp[i][j] == '@') {
					ans = min(ans, disA[id(point(i, j))] + disB[id(point(i, j))]);
				}
			}
		}
		printf("%d
", ans*11);
	}
}

N - Find a way

找一个点到两个点的距离和最小

/*
 * @Author: zhl
 * @LastEditTime: 2020-12-23 10:29:17
 */
#include<cstdio>
#include<cstring>
#include<queue>
#include<bits/stdc++.h>
using namespace std;
#define min(a,b) (a < b ? a : b)
int n, m;
char mp[220][220];
const int dx[] = { 1,0,-1,0 };
const int dy[] = { 0,1,0,-1 };
struct point {
	int x, y;
	point() {}
	point(int _x, int _y) { x = _x, y = _y; }
	bool operator < (const point& b)const {
		if (x != b.x)return x < b.x;
		return y < b.y;
	}
}stA,stB;
int id(point x) {return (x.x - 1) * m + x.y;}
int disA[50000], disB[50000], vis[50000];

void bfs(point st,int* dis) {
	memset(vis, 0, sizeof vis); vis[id(st)] = 1;
	queue<point>Q;Q.push(st);
	int now = 0;
	while (not Q.empty()) {
		int sz = Q.size();
		while (sz--) {
			point tp = Q.front(); Q.pop();
			dis[id(tp)] = now;
			for (int i = 0; i < 4; i++) {
				point nxt = point(tp.x + dx[i], tp.y + dy[i]);
				if (nxt.x < 1 or nxt.x > n or nxt.y < 1 or nxt.y > m or vis[id(nxt)])continue;
				if (mp[nxt.x][nxt.y] == '#')continue;
				vis[id(nxt)] = 1;
				Q.push(nxt);
			}
		}
		now++;
	}
}
int main() {
	while (~scanf("%d%d", &n, &m)) {
		for (int i = 1; i <= n; i++) {
			scanf("%s", mp[i] + 1);
			for (int j = 1; j <= m; j++) {
				if (mp[i][j] == 'M')stA = point(i, j);
				if (mp[i][j] == 'Y')stB = point(i, j);
			}
		}
		memset(disA, 0x3f, sizeof disA); memset(disB, 0x3f, sizeof disB);
		bfs(stA, disA); bfs(stB, disB);
		int ans = 0x3f3f3f3f;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				if (mp[i][j] == '@') {
					ans = min(ans, disA[id(point(i, j))] + disB[id(point(i, j))]);
				}
			}
		}
		printf("%d
", ans*11);
	}
}