ACM刷题之POJ————Pots

Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20922		Accepted: 8916		Special Judge

Description

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
DROP(i) empty the pot i to the drain;
POUR(i,j) pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).

Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

Input

On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).

Output

The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

Sample Input

3 5 4

Sample Output

6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)

Source

Northeastern Europe 2002, Western Subregion

这题是简单的搜索，唯一注意的就是路径打印。

这里我用了字符串来记录路径，感觉挺不错的。

下面是ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<iostream>
using namespace std;
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N=2e5+7;
int n , m , x;

bool vis[200][200];
int val[200][200];


struct nodes{
	int x, y;
	string step;
	int cnt;
};


int bfs() {
	int cnt = 0;
	CLR(vis, 0);
	queue<nodes> q;
	nodes nd1;
	nd1.x = n;
	nd1.y = 0;
	nd1.cnt = 1;
	nd1.step = "FILL(1)\n";
	q.push(nd1);
	
	nd1.x = 0;
	nd1.y = m;
	nd1.step = "FILL(2)\n";
	q.push(nd1);
	
	vis[n][0] = 1;
	vis[0][m] = 1;
	
	while(!q.empty()) {
		nodes nd = q.front();
		q.pop();
		if (nd.x == x || nd.y == x) {
			cnt = nd.cnt;
			cout<<cnt<<"\n";
			cout<<nd.step;
			break;
		}
		++nd.cnt;
		nodes nd2;
		nd2.cnt = nd.cnt;
		if (nd.x < n && !vis[n][nd.y]) {
			nd2.x = n;
			nd2.y = nd.y;
			nd2.step = nd.step + "FILL(1)\n";
			q.push(nd2);
			vis[nd2.x][nd2.y] = 1;
		}
		if (nd.y < m && !vis[nd.x][m]) {
			nd2.x = nd.x;
			nd2.y = m;
			nd2.step = nd.step + "FILL(2)\n";
			q.push(nd2);
			vis[nd2.x][nd2.y] = 1;
		}
		
		if (nd.x > 0 && nd.y != m) {
			if (nd.x >= (m - nd.y) && nd.x != 0) {
				nd2.x = (nd.x - (m - nd.y));
				nd2.y = m;
				
			} else {
				
				nd2.y = (nd.x + nd.y);
				nd2.x = 0;
			}
			if (!vis[nd2.x][nd2.y]) {
				nd2.step = nd.step + "POUR(1,2)\n";
				q.push(nd2);
				vis[nd2.x][nd2.y] = 1;
			}
			
		}
		
		if (nd.y > 0 && nd.x != n) {
			if (nd.y >= (n - nd.x) && nd.y != 0) {
				nd2.y = (nd.y - (n - nd.x));
				nd2.x = n;
			} else {
				
				nd2.x = (nd.y + nd.x);
				nd2.y = 0;
			}
			if (!vis[nd2.x][nd2.y]) {
				nd2.step = nd.step + "POUR(2,1)\n";
				q.push(nd2);
				vis[nd2.x][nd2.y] = 1;	
			}
		}
		
		if (nd.x > 0 && !vis[0][nd.y]) {
			nd2.x = 0;
			nd2.y = nd.y;
			nd2.step = nd.step + "DROP(1)\n";
			q.push(nd2);
			vis[nd2.x][nd2.y] = 1;
		}
		if (nd.y > 0 && !vis[nd.x][0]) {
			nd2.x = nd.x;
			nd2.y = 0;
			nd2.step = nd.step + "DROP(2)\n";
			q.push(nd2);
			vis[nd2.x][nd2.y] = 1;
		}
		
	}
	return cnt;
}

int main()
{
//	freopen("f:/input.txt", "r", stdin);
	while(cin>>n>>m>>x) {
		if (bfs() == 0) {
			cout<<"impossible"<<endl;
		} 
	}
	
}

ACM刷题之POJ————Pots

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