P1379 八数码难题

合集 - 思维题(99)

1.大雪球2024-01-162.E. Final Countdown2024-02-183.D. Divisible Pairs2024-02-194.F. Chat Screenshots2024-02-195.G. One-Dimensional Puzzle2024-02-206.P8775 [蓝桥杯 2022 省 A] 青蛙过河2024-02-217.Watering an Array2024-02-238.Dora and Search2024-02-249.P1137 旅行计划2024-02-2410.P3067 [USACO12OPEN] Balanced Cow Subsets G2024-02-2611.P4799 [CEOI2015 Day2] 世界冰球锦标赛2024-02-2612.A. Brick Wall2024-02-2813.B. Minimize Inversions2024-02-2814.C. XOR-distance2024-02-2815.A. Moving Chips2024-02-2816.B. Monsters Attack!2024-02-2817.C. Find B2024-02-2818.D. Slimes2024-02-2819.C. Turtle Fingers: Count the Values of k2024-02-2820.D. Turtle Tenacity: Continual Mods2024-02-2821.D. Vlad and Division2024-02-2822.C. LR-remainders2024-02-2823.P5322 [BJOI2019] 排兵布阵2024-03-0324.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-0325.P1156 垃圾陷阱2024-03-0326.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-0427.P3047 [USACO12FEB] Nearby Cows G2024-03-0428.P5020 [NOIP2018 提高组] 货币系统2024-03-0429.P3957 [NOIP2017 普及组] 跳房子2024-03-0630.P1550 [USACO08OCT] Watering Hole G2024-03-0631.P1525 [NOIP2010 提高组] 关押罪犯2024-03-07

32.P1379 八数码难题2024-03-07

33.P6121 [USACO16OPEN] Closing the Farm G2024-03-0834.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0835.P9632 [ICPC2020 Nanjing R] K Co-prime Permutation2024-03-0836.P9825 [ICPC2020 Shanghai R] Fibonacci2024-03-0837.P2034 选择数字2024-03-0938.P7072 [CSP-J2020] 直播获奖2024-03-0939.P1714 切蛋糕2024-03-0940.P4147 玉蟾宫2024-03-1041.P2866 [USACO06NOV] Bad Hair Day S2024-03-1042.P2032 扫描2024-03-1043.P1884 [USACO12FEB] Overplanting S2024-03-1044.D. Lonely Mountain Dungeons2024-03-1245.F. Microcycle2024-03-1246.E. Increasing Subsequences__22024-03-1347.D. Valiant's New Map2024-03-1448.Array Repetition2024-03-1449.F. Rudolf and Imbalance2024-03-1450.D. Exam in MAC2024-03-1551.G. Rudolf and Subway2024-03-1552.C. Ehab and Path-etic MEXs2024-03-1753.A. String Transformation 12024-03-1754.D. Solve The Maze2024-03-1855.B. Array Fix2024-03-1856.B. Equal XOR2024-03-1857.F. Maximum White Subtree2024-03-1958.P1652 圆2024-03-2059.P2181 对角线2024-03-2060.P1325 雷达安装2024-03-2061.P3194 [HNOI2008] 水平可见直线2024-03-2062.B - Make Many Triangles2024-03-2363.[ARC172A] Chocolate2024-03-2564.C. Theofanis' Nightmare2024-03-2765.D. Birthday Gift2024-03-2766.C. Arrow Path2024-03-2767.P1265 公路修建2024-03-2868.P1284 三角形牧场2024-03-2869.Kirill and Mushrooms2024-03-2870.C. Sasha and the Casino2024-03-2871.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-0172.P5536 【XR-3】核心城市2024-04-0273.P5836 [USACO19DEC] Milk Visits S2024-04-0274.P4391 [BOI2009] Radio Transmission 无线传输2024-04-0475.P3435 [POI2006] OKR-Periods of Words2024-04-0476.P5960 【模板】差分约束2024-04-0477.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-0578.E. Long Inversions2024-04-0979.C. Deep Down Below2024-04-0980.C. Rings2024-04-0981.F. Unfair Game2024-04-1082.G. GCD on a grid2024-04-1083.C. Inversion Graph2024-04-1084.Increase Subarray Sums2024-04-1085.D. Cross Coloring2024-04-1286.M. Triangle Construction2024-04-1287.D2. Set To Max (Hard Version)2024-04-1288.C. Array Game2024-04-1289.F. Alex's whims2024-04-1290.Queue Sort2024-04-1291.P1908 逆序对2024-04-1392.P5094 [USACO04OPEN] MooFest G 加强版2024-04-1593.P5149 会议座位2024-04-1594.P10337 [UESTCPC 2024] 操作序列2024-04-1595.P10330 [UESTCPC 2024] 黑白珠串2024-04-1596.P10252 线性变换2024-04-1697.P3901 数列找不同2024-04-1698.P10288 [GESP样题 八级] 区间2024-04-1699.P5677 [GZOI2017] 配对统计2024-04-17

收起

原题链接

题解1：朴素广搜

注意细节

code1

#include<bits/stdc++.h>
using namespace std;
int poi[4]={-3,1,-1,3};
int main()
{
    string s;
    cin>>s;
    queue<string> q;
    map<string,int> rec;
    q.push(s);
    rec[s]=1;
    while(q.size())
    {
        string now=q.front();
        q.pop();
        if(now=="123804765")break;
        for(int i=0;i<9;i++)
        if(now[i]=='0')
        {
            for(int j=0;j<4;j++)
            {
                int id=i+poi[j];
                if(id<0||id>8||id==2&&i==3||id==3&&i==2||id==5&&i==6||id==6&&i==5)continue;//注意细节，如果位于矩阵边缘就不能进行某些操作！！
                string next=now;
                swap(next[i],next[id]);
                if((rec[next]>rec[now]+1||!rec[next])&&(rec[next]=rec[now]+1)) q.push(next);
            }
            break;
        }
    }


    cout<<rec["123804765"]-1;
    return 0;
}

题解2

A*搜索，其实在我看来像是一种贪心，通过预估函数预估走这个节点到达终点需要的花费和已经花费的，然后按花费从小到大排序，优先遍历小的
由于本题数据较小，所以这里优先遍历能通过的，再因为本题数据较小，所以预估能到达的都遍历
又因为迭代深度达到一定时就一定可以到达，所以二分深度，然后贪心深搜（因为深度一定，快速达到深度，比广搜更高效）

code2

#include<bits/stdc++.h>
using namespace std;
int xx[4]={1,0,-1,0},yy[4]={0,1,0,-1};
string raw="123804765";
string s;
int dif()
{
    int ans=0;
    for(int i=0;i<9;i++)
    {
        if(s[i]=='0') continue;
        int pos=raw.find(s[i]);
        int x=i/3,y=i%3;
        int x1=pos/3,y1=pos%3;
        ans+=abs(x-x1)+abs(y-y1);
    }
    return ans;
}

int depth;
int ss(int sum,int fa)
{
    int dif1=dif();
    if(dif1==0)return 1;
    if(sum+dif1>depth)return 0;
    int pos=s.find('0');
    int x=pos/3,y=pos%3;;
    for(int i=0;i<4;i++)
    {
        int x1=x+xx[i],y1=y+yy[i];
        if(x1>2||x1<0||y1>2||y1<0) continue;
        int pos1=x1*3+y1;
        if(pos1==fa) continue;
        swap(s[pos],s[pos1]);
        int re=ss(sum+1,pos);
        swap(s[pos],s[pos1]);
        if(re)return 1;
    }
    return 0;
}
int main()
{
    cin>>s;
    int l=-1,r=27;
    while(l<r-1)
    {
        depth=(l+r)/2;
        if(ss(0,-1)) r=depth;
        else l=depth;
    }
    cout<<r;

    return 0;
}