poj 1077 Eight（bfs，dbfs, A*）

代码如下：

bfs:

#include <iostream>
#include <map>
#include <algorithm>
#include <string>
#include <queue>
using namespace std;
typedef long long LL;

int eight[3][3], xx, xy, dx[]={0,0,-1,1,0}, dy[]={0,1,0,0,-1};
char dir[] = "orudl";
LL tar;

LL encode(){
    LL s=eight[2][2];
    for(int i=7; i>=0; --i){
        s <<=4;
        s |= eight[i/3][i%3];
    }
    return s;
}
void decode(LL s){
    for(int i=0; i<9; ++i){
        eight[i/3][i%3] = s%16;
        if(s%16 == 0) 
            xx = i/3, xy = i%3;
        s >>=4;
    }
}
queue<LL> que;
map<LL,pair<int,LL> > mp;// pair:d,ps

void print(LL s){
    int d = mp[tar].first;
    LL ts=s;
    string ope;
    while(d!=-1){
        ope.push_back(dir[d]);
        ts = mp[ts].second;
        d = mp[ts].first;
    }
    reverse(ope.begin(), ope.end());
    cout<< ope<< "\n";
}
void bfs(){
    mp.clear();
    LL s = encode(), s2;
    que.push(s); mp[s] = make_pair(-1,0);
    while(!que.empty()){
        s = que.front(); que.pop();
        for(int i=1; i<5; ++i){
            decode(s);
            int r,c;
            r = xx+dx[i], c = xy+dy[i];
            if(r<0 || c<0 || r==3 || c==3)
                continue;
            swap(eight[xx][xy], eight[r][c]);
            s2 = encode();
            if(mp.find(s2)!=mp.end())
                continue;
            mp[s2] = make_pair(i,s);
            if(s2==tar){
                print(s2);
                return ;
            }
            else que.push(s2);
        }
    }
}
int main(){
    std::ios::sync_with_stdio(false);
    char c;
    for(int i=0; i<3; ++i){
        for(int j=0; j<3; ++j){
            cin>>c;
            if(c=='x')
                eight[i][j]=0, xx = i, xy = j;
            else 
                eight[i][j]= c-'0';
        }
    }
    tar=8;
    for(int i=7; i>0; --i)
        tar<<=4, tar |= i;
    int cnt=0; 
    for(int i=0; i<9; ++i){
        for(int pi=0; pi<i; ++pi)
            if(eight[pi/3][pi%3] && eight[pi/3][pi%3]<eight[i/3][i%3])
                cnt++;
    }
    if(cnt%2)
        printf("unsolvable\n");
    else 
        bfs();
    return 0;
}

 

dbfs：

 

#include <iostream>
#include <map>
#include <algorithm>
#include <string>
#include <queue>
using namespace std;
typedef long long LL;

int eight[3][3], xx, xy, dx[]={0,0,-1,1,0}, dy[]={0,1,0,0,-1};
char dir[] = "orudl";

LL encode(){
    LL s=eight[2][2];
    for(int i=7; i>=0; --i){
        s <<=4;
        s |= eight[i/3][i%3];
    }
    return s;
}
void decode(LL s){
    for(int i=0; i<9; ++i){
        eight[i/3][i%3] = s%16;
        if(s%16 == 0) 
            xx = i/3, xy = i%3;
        s >>=4;
    }
}
queue<LL> que[2];
map<LL,pair<int,LL> > mp[2];// pair:d,ps

void print(LL s){
    int d = mp[0][s].first;
    LL ts=s;
    string ope;
    while(d!=-1){
        ope.push_back(dir[d]);
        ts = mp[0][ts].second;
        d = mp[0][ts].first;
    }
    reverse(ope.begin(), ope.end());
    cout<< ope;
    ope.clear(), d = mp[1][s].first, ts = s;
    while(d!=-1){
        ope.push_back(dir[d]);
        ts = mp[1][ts].second;
        d = mp[1][ts].first;
    }
    cout << ope << "\n";
}
void expand(int idx){
    static int found = 0;
    if(found){
        while(!que[idx].empty()) que[idx].pop();
        return;
    }
    int len = que[idx].size();
    while(len--){
        LL s = que[idx].front(), s2; que[idx].pop();
        for(int i=1; i<5; ++i){
            decode(s);
            int r,c;
            if(idx==0)
                r = xx+dx[i], c = xy+dy[i];
            else 
                r = xx-dx[i], c = xy-dy[i];
            if(r<0 || c<0 || r==3 || c==3)
                continue;
            swap(eight[xx][xy], eight[r][c]);
            s2 = encode();
            if(mp[idx].find(s2)!=mp[idx].end())
                continue;
            mp[idx][s2] = make_pair(i,s);
            if(mp[idx^1].find(s2)!=mp[idx^1].end()){
                print(s2);
                found=1;
                return ;
            }
            else que[idx].push(s2);
        }
    }
}
void dbfs(){
    mp[0].clear(), mp[1].clear();
    LL tar=8;
    for(int i=7; i>0; --i)
        tar<<=4, tar |= i;
    que[1].push(tar); mp[1][tar]=make_pair(-1,0);// 两者相等??
    tar = encode();
    que[0].push(tar); mp[0][tar] = make_pair(-1,0);
    while(!que[0].empty() && !que[1].empty()){
        if(que[0].size()<que[1].size())
            expand(0);
        else expand(1);
    }
    while(!que[0].empty())
        expand(0);
    while(!que[1].empty())
        expand(1);
}
int main(){
    freopen("C:\\Users\\yyf\\Documents\\CppFiles\\in.txt", "r", stdin);
    std::ios::sync_with_stdio(false);
    char c;
    for(int i=0; i<3; ++i){
        for(int j=0; j<3; ++j){
            cin>>c;
            if(c=='x')
                eight[i][j]=0, xx = i, xy = j;
            else 
                eight[i][j]= c-'0';
        }
    }
    int cnt=0; 
    for(int i=0; i<9; ++i){
        for(int pi=0; pi<i; ++pi)
            if(eight[pi/3][pi%3] && eight[pi/3][pi%3]<eight[i/3][i%3])
                cnt++;
    }
    if(cnt%2)
        printf("unsolvable\n");
    else 
        dbfs();
    return 0;
}

 

 

 A*:

#include <iostream>
#include <map>
#include <algorithm>
#include <string>
#include <queue>
#include <set>
using namespace std;
typedef long long LL;

int eight[3][3], xx, xy, dx[]={0,-1,1,0}, dy[]={1,0,0,-1};
char dir[] = "rudl";
LL init, tar;

LL encode(){
    LL s=eight[2][2];
    for(int i=7; i>=0; --i){
        s <<=4;
        s |= eight[i/3][i%3];
    }
    return s;
}
void decode(LL s){
    for(int i=0; i<9; ++i){
        eight[i/3][i%3] = s%16;
        if(s%16 == 0) 
            xx = i/3, xy = i%3;
        s >>=4;
    }
}
class Node {
public:
    LL s, ps;
    int f,g,h,d;
    Node (LL _s=0, LL _ps=0, int _d=-1, int _f=0, int _g=0, int _h=0){
        s = _s, ps = _ps, f = _f, g = _g, h = _h, d = _d;
    }
    int operator<(const Node & b) const {
        return this->f < b.f || (this->f==b.f && this->s<b.s);
    }
};

multiset<Node> closed, open;
map<LL, multiset<Node>::iterator> inOpen, inClosed;

void print(Node nd){
    int d = nd.d;
    LL ts = nd.s;
    string ope;
    while(d!=-1){
        ope.push_back(dir[d]);
        auto itc = inClosed[nd.ps];
        nd = *itc;
        d = nd.d;
    }
    reverse(ope.begin(), ope.end());
    cout<< ope<< "\n";
}
int hn(LL s){
    decode(s);
    int cnt=0;
    for(int i=0; i<9; ++i){
        if(eight[i/3][i%3] != (i+1)%9) cnt++;
    }
    return cnt;
}
void astar(){
    init = encode();
    int inith = hn(init);
    Node start = Node(init, 0, -1, inith, 0, inith);
    open.insert(start);
    while(!open.empty()){
        Node tn = *open.begin(); inOpen.erase(tn.s); open.erase(tn);
        LL ts = tn.s, s2;
        int tf = tn.f, tg = tn.g, th = tn.h;
        for(int i=0; i<4; ++i){
            decode(ts);
            int r = xx+dx[i], c=xy+dy[i];
            if(r<0 || c<0 || r==3 || c==3) continue;
            swap(eight[r][c], eight[xx][xy]);
            s2 = encode();
            int g = tg+1, h = hn(s2), f = g+h;
            Node nd = Node(s2, ts, i, f, g, h);
            if(s2 == tar){
                inClosed[tn.s]= closed.insert(tn);
                inClosed[nd.s]= closed.insert(nd);
                print(nd);
                return;
            }
            auto ito = inOpen.find(s2), itc = inClosed.find(s2);
            if(ito == inOpen.end() && itc == inClosed.end()){
                inOpen[nd.s]=open.insert(nd);
            }
            else if(ito == inOpen.end() && f < itc->second->f){
                closed.erase(itc->second); inClosed.erase(itc); 
                inOpen[nd.s]=open.insert(nd);
            }
            else if(itc == inClosed.end() && f < ito->second->f){
                // que 里有相同s不同f的node
                open.erase(ito->second), inOpen[nd.s]=open.insert(nd);
            }
        }
        inClosed[tn.s] = closed.insert(tn);
    }
}
int main(){
    std::ios::sync_with_stdio(false);
    char c;
    for(int i=0; i<3; ++i){
        for(int j=0; j<3; ++j){
            cin>>c;
            if(c=='x')
                eight[i][j]=0, xx = i, xy = j;
            else 
                eight[i][j]= c-'0';
        }
    }
    tar=8;
    for(int i=7; i>0; --i)
        tar<<=4, tar |= i;
    int cnt=0; 
    for(int i=0; i<9; ++i){
        for(int pi=0; pi<i; ++pi)
            if(eight[pi/3][pi%3] && eight[pi/3][pi%3]<eight[i/3][i%3])
                cnt++;
    }
    if(cnt%2)
        printf("unsolvable\n");
    else 
        astar();
    return 0;
}