ZOJ 1005 - Jugs | UVA 571 - Jugs-优快云博客

本文探讨了一种广搜算法在解决特定问题时的状态控制，通过实例展示了算法的运用和优化策略。

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广搜水过去了,状态很少.

#include <iostream>
#include <memory.h>
#include <cstdio>
#include <map>
using namespace std;
const int MAX = 10001001;
const char * ins[] ={"fill A", "fill B", "empty A", "empty B", "pour A B", "pour B A"};

struct Node{
	int v, i, fa;
	Node(int v_ = 0, int i_ = 0, int fa_ = -1):v(v_), i(i_), fa(fa_){}
}q[1000];

map<int, bool> vis;
int CA, CB, N;

inline void put_into_queue(int A, int B, int f, int & r, int i){
	Node n;
	n.v = A * 10000 + B;
	n.i = i;
	n.fa = f;
	if(!vis[n.v]){
		q[r++] = n;
		vis[n.v] = 1;
	}
}

int bfs(){
	//memset(vis, 0, sizeof(vis));
	vis.clear();
	int f = 0, r = 0;
	q[r++] = Node(0, 0, -1);
	vis[0] = 1;
	while(f < r){
		Node n = q[f];
		
		int A = n.v / 10000;
		int B = n.v % 10000;
		if(B == N)return f;
		//fill A
		put_into_queue(CA, B, f, r, 0);
		//fill B
		put_into_queue(A, CB, f, r, 1);
		//empty A
		put_into_queue(0, B, f, r, 2);
		//empty B
		put_into_queue(A, 0, f, r, 3);
		//pour A into B
		put_into_queue(A - min(A, CB - B), B + min(A, CB - B), f, r, 4);
		//pour B into A
		put_into_queue(A + min(B, CA - A), B - min(B, CA - A), f, r, 5);
		++f;
	}
}

void print_path(int i){
	if(i == 0)return;
	print_path(q[i].fa);
	printf("%s\n", ins[q[i].i]);
}

int main(int argc, char const *argv[]){
	while(scanf("%d%d%d", &CA, &CB, &N) == 3){
		int p = bfs();
		print_path(p);
		printf("success\n");
	}
	return 0;
}