2024年3月20题解

M - A计划

解题思路

1. 从起点出发进行bfs，把可以走的位置入队
2. 但碰到了传送阵的话，需要判断是不是墙和传送阵，因为这两种情况都会让骑士找不到公主
3. 剩下的就是正常bfs的过程

代码实现

#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <stdbool.h>
#include <limits.h>
#define u unsigned
#define ll long long
#define sc scanf
#define pr printf 
#define fr(i, j, n) for (int i = j; i < n; i++)
#define N 10


typedef struct {
	int x;
	int y;
	int z;
	int s;
}p;

int d[4][3] = {
	{0, -1, 0}, {0, 1, 0}, {0, 0, -1}, {0, 0, 1}
};
int t;
int n;
int m;
int c;
char s[2][N][N];

int bfs();
int min1(int a, int b) {
	return a <= b ? a : b;
}

int main(int argc, char* argv[])
{
	sc("%d", &c);

	while (c--) {
		sc("%d%d%d", &n, &m, &t);

		for (int i = 0; i < 2; i++) {
			for (int j = 0; j < n; j++) {
				sc("%s", s[i][j]);
			}
		}

		if (bfs() <= t) {
			pr("YES\n");
		}
		else {
			pr("NO\n");
		}
	}

	return 0;
}

int bfs()
{
	p q[N * N * 2] = { 0 };
	bool v[2][N][N] = { 0 };
	p t = { 0 };
	int ans = INT_MAX;
	int l = 0;
	int r = 0;
	p temp = { 0 };

	q[r++] = t;
	v[0][0][0] = 1;

	while (l < r) {
		t = q[l++];

		if (s[t.x][t.y][t.z] == '#')
		{
			t.x = !t.x;
			if (s[t.x][t.y][t.z] == '*') {
				continue;
			}
			if (s[t.x][t.y][t.z] == '#') {//坑点1:和洛谷的那个题目不一样，这个传送过去可能是#，那么就一直横跳
				continue;
			}
		}
		if (s[t.x][t.y][t.z] == 'P') {//传送的地方还可能是终点所以也要判断
			ans = min1(ans, t.s);
			continue;
		}

		for (int i = 0; i < 4; i++) {
			int nx = t.x + d[i][0];
			int ny = t.y + d[i][1];
			int nz = t.z + d[i][2];

			if (nx >= 0 && nx <= 1 && ny >= 0 && ny < n && nz >= 0 && nz < m && s[nx][ny][nz] != '*' && !v[nx][ny][nz]) {
				temp.x = nx;
				temp.y = ny;
				temp.z = nz;
				temp.s = t.s + 1;
				v[nx][ny][nz] = 1;
				q[r++] = temp;
			}
		}
	}

	return ans;
}

Pots

解题思路

1. 因为是求最短序列，所以用bfs来搜索答案
2. bfs来遍历6中情况，因为有两个瓶子，一共三种操作

代码实现

#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <stdbool.h>
#include <limits.h>
#define u unsigned
#define ll long long
#define sc scanf
#define pr printf 
#define fr(i, j, n) for (int i = j; i < n; i++)
#define N 101

typedef struct {
	int a;
	int b;
	int path[200];
	int size;
}p;

int a;
int b;
int c;
bool is;
bool v[N][N][7];
int ans = INT_MAX;
p ans1 = { 0 };
p q[N * N * 6] = { 0 };

int min1(int a, int b) {
	return a <= b ? a : b;
}

void bfs()
{
	p t = { 0 };
	p temp = { 0 };
	int l = 0;
	int r = 0;

	q[r++] = t;

	while (l < r) {
		t = q[l++];


		if (t.a == c || t.b == c) {
			pr("%d\n", t.size);
			for (int i = 0; i < t.size; i++) {
				switch (t.path[i]) {
				case 1: {
					pr("FILL(1)\n");
					break;
				}
				case 2:
				{
					pr("FILL(2)\n");
					break;
				}
				case 3: {
					pr("DROP(1)\n");
					break;
				}
				case 4:
				{
					pr("DROP(2)\n");
					break;
				}
				case 5:
				{
					pr("POUR(1,2)\n");
					break;
				}
				case 6: {
					pr("POUR(2,1)\n");
					break;
				}
				}
			}
			is = 1;
			break;
		}

		for (int i = 1; i <= 6; i++) {
			temp = { 0 };
			switch (i) {
			case 1: {
				if (t.a != a && !v[t.a][t.b][i]) {
					v[t.a][t.b][i] = 1;
					temp = t;
					temp.a = a;
					temp.path[temp.size] = 1;
					temp.size++;
					q[r++] = temp;
				}
				break;
			}
			case 2: {
				if (t.b != b && !v[t.a][t.b][i]) {
					v[t.a][t.b][i] = 1;
					temp = t;
					temp.b = b;
					temp.path[temp.size] = 2;
					temp.size++;
					q[r++] = temp;
				}
				break;
			}
			case 3: {
				if (t.a != 0 && !v[t.a][t.b][i]) {
					v[t.a][t.b][i] = 1;
					temp = t;
					temp.a = 0;
					temp.path[temp.size] = 3;
					temp.size++;
					q[r++] = temp;
				}
				break;
			}
			case 4: {
				if (t.b != 0 && !v[t.a][t.b][i]) {
					v[t.a][t.b][i] = 1;
					temp = t;
					temp.b = 0;
					temp.path[temp.size] = 4;
					temp.size++;
					q[r++] = temp;
				}
				break;
			}
			case 5: {
				if (t.b != b && t.a != 0 && !v[t.a][t.b][i]) {
					v[t.a][t.b][i] = 1;
					temp = t;
					if (temp.b + temp.a >= b) {
						temp.a -= b - temp.b;
						temp.b = b;
					}
					else {
						temp.b += temp.a;
						temp.a = 0;
					}
					temp.path[temp.size] = 5;
					temp.size++;
					q[r++] = temp;
				}
				break;
			}
			case 6: {
				if (t.a != a && t.b != 0 && !v[t.a][t.b][i]) {
					v[t.a][t.b][i] = 1;
					temp = t;
					if (temp.b + temp.a >= a) {
						temp.b -= a - temp.a;
						temp.a = a;
					}
					else {
						temp.a += temp.b;
						temp.b = 0;
					}
					temp.path[temp.size] = 6;
					temp.size++;
					q[r++] = temp;
				}
				break;
			}
			}
		}
	}

	if (!is) {
		pr("impossible\n");
	}

}

int main(int argc, char* argv[])
{
	sc("%d%d%d", &a, &b, &c);

	bfs();

	return 0;
}