搜索小练#8

题目

这里很容易想到广搜，设初始状态为\((i,j)\)，一共就只有六种变化

A杯倒满，B杯倒满，A杯倒出完，B杯倒出完，A到给B，B到给A

任意一种状态\(i\) 或者 \(j\)达到 \(C\) 就停止，注意要打印路径，我这里用一个string存的操作顺序。

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

const int N = 1e2+100;

int a, b, c;
int v[N][N];
bool flag = false;
string str[10] = {"", "FILL(1)", "FILL(2)", "DROP(1)", "DROP(2)", "POUR(1,2)","POUR(2,1)"};

struct node {
	int x, y, step; string s;
	node (int x, int y, int step, string s):x(x), y(y), step(step),s(s){}
};

inline void bfs () {
	queue<node>q;
	v[0][0] = 1;
	q.push(node(0, 0, 0, "0"));
	while(!q.empty()) {
	    /* code */
		node tmp = q.front(); q.pop();
		if (tmp.x == c || tmp.y == c) {
			flag = 1;
			cout << tmp.step << '\n';
			for (int i = 1; i < tmp.s.length(); ++ i)
				cout << str[tmp.s[i]-'0'] << '\n';
			return;
		}
		if (tmp.x < a)
			if (!v[a][tmp.y])
				v[a][tmp.y] = 1, q.push(node(a, tmp.y, tmp.step+1, tmp.s+"1"));
		if (tmp.y < b)
			if (!v[tmp.x][b])
				v[tmp.x][b] = 1, q.push(node(tmp.x, b, tmp.step+1, tmp.s+"2"));
		if (tmp.x != 0)
			if (!v[0][tmp.y])
				v[0][tmp.y] = 1, q.push(node(0, tmp.y, tmp.step+1, tmp.s+"3"));
		if (tmp.y != 0)
			if (!v[tmp.x][0])
				v[tmp.x][0] = 1, q.push(node(tmp.x, 0, tmp.step+1, tmp.s+"4"));
		if (tmp.x != 0 && tmp.y < b) {
			int nx, ny;
			if (tmp.x <= b-tmp.y) nx = 0, ny = tmp.x+tmp.y;
			else nx = tmp.x+tmp.y-b, ny = b;
			if (!v[nx][ny]) 
				v[nx][ny] = 1, q.push(node(nx, ny, tmp.step+1, tmp.s+"5"));
		}
		if (tmp.y != 0 && tmp.x < a) {
			int nx, ny;
			if (tmp.y <= a-tmp.x) nx = tmp.x+tmp.y, ny = 0;
			else nx = a, ny = tmp.x+tmp.y-a;
			if (!v[nx][ny]) 
				v[nx][ny] = 1, q.push(node(nx, ny, tmp.step+1, tmp.s+"6"));
		}
	}
}

int main () {
	cin >> a >> b >> c;
	bfs ();
 	if (!flag) puts ("impossible");	
	return 0;
}