2017多校第10场 HDU 6171 Admiral 双向BFS或者A*搜索

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6171

题意： 给你一个高度为6的塔形数组，你每次只能将0与他上下相邻的某个数交换，问最少交换多少次可以变为初始状态，若需要的步数大于20，直接输出too difficult，初始状态为：

0 
1 1 
2 2 2 
3 3 3 3 
4 4 4 4 4 
5 5 5 5 5 5

解法：两种方法，一种是双向BFS+Hash，另外是A*估价+Hash。

双向搜索参考：http://blog.csdn.net/cillyb/article/details/77587228

 

//双BFS
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
const int dir[4][2] = {{-1,-1},{-1,0},{1,0},{1,1}};
typedef unsigned long long uLL;
struct node{
    int val[6][6];
    int r, c, k, step;
};
map <uLL, int> book[2];
uLL Hash(node x){
    uLL ret = 0;
    for(int i=0; i<6; i++)
        for(int j=0; j<=i; j++)
            ret = ret*6+x.val[i][j];
    return ret;
}
int BFS(node s, node e){
    queue <node> q;
    book[0].clear();
    book[1].clear();
    s.k = 0;
    e.k = 1;
    s.step = e.step = 0;
    book[s.k][Hash(s)] = 0;
    book[e.k][Hash(e)] = 0;
    q.push(s);
    q.push(e);
    while(q.size()){
        node u = q.front(); q.pop();
        uLL tmp = Hash(u);
        if(book[!u.k].count(tmp)){
            if(book[!u.k][tmp]+u.step<=20){
                return book[!u.k][tmp]+u.step;
            }
            else continue;
        }
        if(u.step >= 10) continue;
        for(int i=0; i<4; i++){
            node t = u;
            t.r += dir[i][0];
            t.c += dir[i][1];
            if(t.r >= 6 || t.r < 0 || t.c > t.r || t.c < 0) continue;
            swap(t.val[t.r][t.c], t.val[u.r][u.c]);
            tmp = Hash(t);
            if(book[t.k].count(tmp)) continue;
            t.step++;
            book[t.k][tmp] = t.step;
            q.push(t);
        }
    }
    return -1;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        node s,e;
        e.r = e.c = 0;
        for(int i=0; i<6; i++){
            for(int j=0; j<=i; j++){
                scanf("%d", &s.val[i][j]);
                if(!s.val[i][j])
                    s.r = i, s.c = j;
                e.val[i][j] = i;
            }
        }
        int ans = BFS(s, e);
        if(ans == -1) puts("too difficult");
        else printf("%d\n", ans);
    }
    return 0;
}

 

放一个队友的A*搜索：

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long  uLL;
int lowPos[6][6];
pair<int, int> ppPos[22];
const int dir[4][2] = {{-1, -1}, {-1, 0}, {1, 0}, {1, 1}};
void init()
{
    int jsq = 0;
    int now = 0;
    for(int i = 0; i < 6; i++)
        for(int j = 0; j <= i; j++)
        {
            lowPos[i][j] = now;
            ppPos[jsq] = make_pair(i, j);
            jsq ++;
            now += 3;
        }
}
void setStatus(uLL &status, int x, int y, uLL val)
{
    int pos = lowPos[x][y];
    status &= ~(1ULL << pos);
    status &= ~(1ULL << pos + 1);
    status &= ~(1ULL << pos + 2);
    status |= val << pos;
    return;
}
void swapStatus(uLL &status, int a, int b, int x, int y)
{
    int pos1 = lowPos[a][b];
    int pos2 = lowPos[x][y];
    uLL val1 = status >> pos1 & 7;
    uLL val2 = status >> pos2 & 7;
    status &= ~(1ULL << pos1);
    status &= ~(1ULL << pos1 + 1);
    status &= ~(1ULL << pos1 + 2);
    status &= ~(1ULL << pos2);
    status &= ~(1ULL << pos2 + 1);
    status &= ~(1ULL << pos2 + 2);
    status |= val1 << pos2;
    status |= val2 << pos1;
}
pair<int, int> getZeroPos(uLL status)
{
    for(int i = 0; i < 21; i++)
    {
        uLL val = status & 7;
        if(val == 0)
            return ppPos[i];
        status >>= 3;
    }
}
int Compare(uLL endStatus, uLL nowStatus)
{
    int ret = 0;
    for(int i = 0; i < 21; i++)
    {
        uLL val1 = endStatus & 7;
        uLL val2 = nowStatus & 7;
        if(val1 != val2)
            ret++;
        endStatus >>= 3;
        nowStatus >>= 3;
    }
    return ret;
}
int main()
{
    init();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        uLL startStatus = 0;
        for(int i = 0; i < 6; i++)
            for(int j = 0; j <= i; j++)
            {
                int x;
                scanf("%d", &x);
                setStatus(startStatus, i, j, x);
            }
        uLL endStatus = 0;
        for(int i = 0; i < 6; i++)
            for(int j = 0; j <= i; j++)
                setStatus(endStatus, i, j, i);
        queue<uLL> q;
        unordered_map<uLL, int> mp;
        mp[startStatus] = 0;
        q.emplace(startStatus);
        while(!q.empty())
        {
            uLL u = q.front();
            q.pop();
            if(u == endStatus)
                break;
            int d = mp[u];
            if(d == 20)
                break;
            pair<int, int> zpos = getZeroPos(u);
            int i = zpos.first;
            int j = zpos.second;
            for(int k = 0; k < 4; k++)
            {
                int x = i + dir[k][0];
                int y = j + dir[k][1];
                if(x < 0 || x > 5 || y < 0 || y > x)
                    continue;
                uLL newStatus = u;
                swapStatus(newStatus, i, j, x, y);
                if(mp.count(newStatus))
                    continue;
                mp[newStatus] = d + 1;
                int famly = Compare(endStatus, newStatus);
                if(famly + d + 2 > 20)
                    continue;
                q.emplace(newStatus);
            }
        }
        if(mp.count(endStatus) == 0)
            puts("too difficult");
        else
            printf("%d\n", mp[endStatus]);
    }
    return 0;
}