翻纸牌游戏_hdu_2209(双向广搜).java

翻纸牌游戏

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1885    Accepted Submission(s): 661


Problem Description

有一种纸牌游戏，很有意思，给你N张纸牌，一字排开，纸牌有正反两面，开始的纸牌可能是一种乱的状态（有些朝正，有些朝反），现在你需要整理这些纸牌。但是麻烦的是，每当你翻一张纸牌（由正翻到反，或者有反翻到正）时，他左右两张纸牌（最左边和最右边的纸牌，只会影响附近一张）也必须跟着翻动，现在给你一个乱的状态，问你能否把他们整理好，使得每张纸牌都正面朝上，如果可以，最少需要多少次操作。

 

Input

有多个case，每个case输入一行01符号串（长度不超过20），1表示反面朝上，0表示正面朝上。

 

Output

对于每组case，如果可以翻，输出最少需要翻动的次数，否则输出NO。

 

Sample Input

01
011

 

Sample Output

NO
1

 

Author

wangye

 

Recommend

wangye   |   We have carefully selected several similar problems for you:  2102 1495 1180 2201 2200 

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Queue;

public class Main {//超时
static String sf="";
static int len=0;
//static HashMap<StringBuffer,Integer> map;
static boolean ok=false;
public static void main(String[] args) {
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
String s;
try {
while((s=bf.readLine())!=null){
ok=false;
StringBuffer ss=new StringBuffer(s);
len=ss.length();
sf="";
for(int i=0;i<len;i++)
sf+="0";
HashMap<String,Integer> map=new HashMap<String, Integer>();
map.put(s, 0);
P p=new P();
p.s=ss;
p.num=0;
Queue<P> q=new LinkedList<P>();
q.add(p);
bfs(q,map);
if(!ok)
System.out.println("No");
}
} catch (Exception e) {
e.printStackTrace();
}

}

private static void bfs(Queue<P> q,HashMap<String,Integer> map) {

while(!q.isEmpty()){
boolean okk=false;
if(okk){
ok=true;
return;

}
P d=new P();
d=q.poll();
String k=d.s.toString();
if(k.equals(sf)){
System.out.println(d.num);
ok=true;
return;
}
StringBuffer g=d.s;
for(int i=0;i<len;i++){
StringBuffer f=new StringBuffer(g);
if(f.charAt(i)=='0'){
f.deleteCharAt(i);
f.insert(i, 1);
}
else{
f.deleteCharAt(i);
f.insert(i, 0);
}
if(i>0){
if(f.charAt(i-1)=='0'){
f.deleteCharAt(i-1);
f.insert(i-1, 1);
}
else{
f.deleteCharAt(i-1);
f.insert(i-1, 0);
}
}
if(i+1<len){
if(f.charAt(i+1)=='0'){
f.deleteCharAt(i+1);
f.insert(i+1, 1);
}
else{
f.deleteCharAt(i+1);
f.insert(i+1, 0);
}
}
String f1=f.toString();
if(map.get(f1)==null){
//	System.out.println("*****"+f);
//	System.out.println(map.get(f));
if(f1.equals(sf)){
System.out.println(d.num+1);
okk=true;
ok=true;
return;
}
P h=new P();
h.s=f;
h.num=d.num+1;
map.put(f1,0);
q.add(h);
}
}

/*if(f.charAt(d.i)=='0'){
f.deleteCharAt(d.i);
f.insert(d.i, 1);
}
else{
f.deleteCharAt(d.i);
f.insert(d.i, 0);
}
if(d.i>0){
if(f.charAt(d.i-1)=='0'){
f.deleteCharAt(d.i-1);
f.insert(d.i-1, 1);
}
else{
f.deleteCharAt(d.i-1);
f.insert(d.i-1, 0);
}
}
if(d.i+1<len){
if(f.charAt(d.i+1)=='0'){
f.deleteCharAt(d.i+1);
f.insert(d.i+1, 1);
}
else{
f.deleteCharAt(d.i+1);
f.insert(d.i+1, 0);
}
}
if(d.i+1!=len){
P g=new P();
g.s=f;
g.i=d.i+1;
g.num=d.num+1;

}*/
}
}
}

class P{
StringBuffer s=null;
int num=0;
}

/*
* 01543093	2013-11-10 12:19:26	Accepted	1004	5406 MS	26468 KB	Java	zhangyi
9547285	2013-11-10 14:27:34	Accepted	2209	5593MS	26476K	1594 B	Java	zhangyi
*/
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.LinkedList;
import java.util.Queue;

public class Main {//AC
static char str[] = new char[30];
static int dis[] = new int[1 << 20], vis[] = new int[1 << 20],
base[] = new int[20], len, aim;

public static void main(String[] args) {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
int i;
aim = 0;
String s = null;
try {
while ((s = bf.readLine()) != null) {
str = s.toCharArray();
len = s.length();
base[0] = 3;
base[len - 1] = 3 << (len - 2);
for (i = 1; i < len - 1; i++)
base[i] = 7 << (i - 1);
if (len == 1) {
if (str[0] == '1')
System.out.println(1);
else
System.out.println(0);
continue;
}
int ans = bfs();
if (ans == -1)
System.out.println("NO");
else
System.out.println(ans);
}
} catch (Exception e) {
e.printStackTrace();
}
}

public static int bfs() {
int i, limit = 1 << len;
Integer now, next;
Queue<Integer> q = new LinkedList<Integer>();

now = 0;
for (i = len - 1; i >= 0; i--)
now = now * 2 + str[i] - '0';
for (i = 0; i < limit; i++)
vis[i] = 0;
dis[now] = 0;
vis[now] = 1;
q.add(now);

while (!q.isEmpty()) {
now = q.poll();
if (now == aim)
return dis[aim];
for (i = 0; i < len; i++) {
next = now ^ base[i];
if (vis[next] != 0)
continue;
dis[next] = dis[now] + 1;
vis[next] = 1;
q.add(next);
}
}

return -1;
}
}