HDU 1043——八数码的多种解题思路（持续更新）
八数码算是一个比较经典的搜索问题了。
hdu 1043： http://acm.hdu.edu.cn/showproblem.php?pid=1043
poj 1077： http://poj.org/problem?id=1077
1.朴素广搜+哈希：
#include
#include
#include
#include
#include
#include
using namespace std;
struct Node {
int s[9];
int loc;
int status;
};
queue q;
struct Path {
int fa;
char d;
};
Path path[500000];
string s,_ans;
int a[10],F[10];
int jud[500000];
int dir_jud[4][9] = {
//rlud
1,1,0,1,1,0,1,1,0,
0,1,1,0,1,1,0,1,1,
0,0,0,1,1,1,1,1,1,
1,1,1,1,1,1,0,0,0
};
int dir[4] ={1,-1,-3,3};
char zifu[10] = {"rlud"};
int Cantor(int a[]) {
int ans = 0;
for (int i = 0; i < 9; i++) {
int t = 0;
for (int j = 0; j < i; j++) {
if(a[j]='1'&&s[i]<='9') {
a[j++] = s[i]-'0';
}
else if(s[i]=='x') {
a[j++] = 9;
}
}
}
int check(int a[]) {
int ans = 0;
for (int i = 0; i < 9; i++) {
for (int j = 0; j < i; j++) {
if(a[i] ();
memset(path,0,sizeof(path));
memset(jud,0,sizeof(jud));
read();cal();
if(check(a)%2) {
puts("unsolvable");
continue;
}
Node start;
for(int i = 0; i < 9; i++) {
start.s[i] = a[i];
if(a[i]==9)start.loc = i;
}
start.status = Cantor(start.s);
q.push(start);
path[start.status].fa = -1;
jud[start.status] = 1;
if(!bfs()) {
puts("unsolvable");
continue;
}
int i = 0;
while (path[i].fa!=-1) {
_ans.push_back(path[i].d);
i = path[i].fa;
}
reverse(_ans.begin(),_ans.end());
cout << _ans << endl;
}
return 0;
} #include
#include
#include
#include
#include
#include
#define MAXN 362880
using namespace std;
int a[10];
char s[1000];
int F[10];
struct Node{
int s[9];
int zero;
int status;
};
queue q;
struct Path{
int fa;
char d;
};
Path path[MAXN];
bool jud[MAXN];
//rlud
char z[10] = {"lrdu"};
int dir[4] = {1,-1,-3,3};
bool pre_d[4][9] = {
1,1,0,1,1,0,1,1,0,
0,1,1,0,1,1,0,1,1,
0,0,0,1,1,1,1,1,1,
1,1,1,1,1,1,0,0,0
};
int Cantor(int c[]) {
int ans = 0;
for (int i = 0;i < 9; i++) {
int t = 0;
for (int j = 0; j < i; j++) {
if (c[j]='1') a[j++] = s[i]-'0';
else if (s[i]=='x') a[j++] = 9;
}
int ans = Cantor(a);
if(jud[ans]==0) {
puts("unsolvable");
}
else{
while(path[ans].fa!=-1) {
printf("%c",path[ans].d);
ans=path[ans].fa;
}
printf("\n");
}
}
return 0;
} 3.双向广搜
// #include
#include
#include
#include
#include
#include
#include
using namespace std;
char s[1000];
string ANS[2];
int f[10];
int a[10];
char z[2][10] = {"lrud","rldu"};
int dir[4] = {1,-1,3,-3};
int p_dir[4][9] = {
1,1,0,1,1,0,1,1,0,
0,1,1,0,1,1,0,1,1,
1,1,1,1,1,1,0,0,0,
0,0,0,1,1,1,1,1,1
};
struct Node {
int zero, status;
int s[9];
Node(){};
Node(int Z,int S){
zero = Z, status = S;
}
};
struct Path {
int fa; char c;
Path(){};
Path(int FA,int C) {
fa = FA, c = C;
}
};
Path path[400000];
queue q[2];
int jud[400000];
int Cantor(int a[]) {
int ans = 0;
for (int i = 0; i < 9; i++) {
int t = 0;
for (int j = 0; j < i; j++)
if (a[i]>a[j]) t++;
ans+=f[9-i-1]*(a[i]-1-t);
}
return ans;
}
int ca;
void dbfs(){
while (!q[0].empty()&&!q[1].empty()) {
for (int j = 0; j < 2; j++) {
for (int i = 0; i < 4; i++) {
if (p_dir[i][q[j].front().zero]) {
Node t = q[j].front();
swap(t.s[t.zero],t.s[t.zero+dir[i]]);
t.status = Cantor(t.s);
if (!jud[t.status]) {
path[t.status].fa = q[j].front().status;
path[t.status].c = z[j][i];
jud[t.status] = j+1;
t.zero += dir[i];
q[j].push(t);
}
else{
if (jud[t.status]==2-j) {
int t1 = q[j].front().status;
int t2 = t.status;
ANS[j]+=z[j][i];
while (path[t1].fa!=-1) {
ANS[j]+=path[t1].c;
t1 = path[t1].fa;
}
while (path[t2].fa!=-1) {
ANS[1-j]+=path[t2].c;
t2 = path[t2].fa;
}
reverse(ANS[1].begin(),ANS[1].end());
cout << ANS[1]<();
ANS[0].clear();ANS[1].clear();
int j = 0, t = 0;
for (int i = 0; i < strlen(s); ++i) {
if (s[i]<='8'&&s[i]>='1') {
a[j++] = s[i]-'0';
}
else if (s[i]=='x'){
a[j++] = 9;
t = j-1;
}
}
if(check(a)%2) {
puts("unsolvable");
continue;
}
int ans = Cantor(a);
q[1].push(Node(t,ans));
q[0].push(Node(8,0));
for (int i = 0 ; i < 9; i++) {
q[0].front().s[i] = i+1;
q[1].front().s[i] = a[i];
}
path[0].fa = path[ans].fa= -1;
jud[0] = 1;
jud[ans] = 2;
dbfs();
}
return 0;
} 注：这题数据好像挺水的，poj无unslovable数据，另外以提交结果看，双向广搜并不快（才不是写的蠢）