启发式搜索：A*与IDEA* 16格拼图为例

题目描述为：

The goal of the 15 puzzle problem is to complete pieces on

4×4

cells where one of the cells is empty space.

In this problem, the space is represented by 0 and pieces are represented by integers from 1 to 15 as shown below.

1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15

You can move a piece toward the empty space at one step. Your goal is to make the pieces the following configuration in the shortest move (fewest steps).

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 0

Write a program which reads an initial state of the puzzle and prints the fewest steps to solve the puzzle.

Input

The 4×4

integers denoting the pieces or space are given.

Output

Print the fewest steps in a line.

Constraints

• The given puzzle is solvable in at most 45 steps.

Sample Input

1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15

Sample Output

8

题目大意:

​ 存在一个4*4的正方形方格图，其中放有0-15的数字，每次只能移动0，求最少多少步可以得到1-15 的顺序

由于普通的bfs时间和空间复杂度较大，因此这里要用到启发式搜索来减少复杂度

IDEA*：通过限制深度的dfs来求得最短

A*：通过估值函数和优先队列每次贪心地选择局部最短来搜索

IDAE*:

#include<bits/stdc++.h>
#define N 4
#define N2 16
#define LIMIT 100
using namespace std;
const int dx[4] = {0,-1,0,1};
const int dy[4] = {1,0,-1,0};
const char dir[4] = {'r','u','l','d'};
int MDT[N2][N2];
struct Puzzle{
    int f[N2],space,MD;
};
Puzzle state;
int limit;
int path[LIMIT];
int getAllMd(Puzzle pz) {
    int sum = 0;
    for(int i = 0;i < N2;i++) {
        if(pz.f[i] == N2) continue;//空格是不计入曼哈顿距离的
        sum += MDT[i][pz.f[i] - 1];
    }
    return sum;
}
bool IsSolved() {
    for(int i = 0;i < N2;i++) if(state.f[i] != i+1) return false;
    return true;
}
bool dfs(int depth,int pre) {
    if(state.MD == 0) return true;
    if(depth + state.MD > limit) return false;
    int sx = state.space / N;
    int sy = state.space % N;
    Puzzle temp;
    for(int r = 0;r < 4;r++) {
        int tx = sx + dx[r];
        int ty = sy + dy[r];
        if(tx < 0 || ty < 0 || tx >= N || ty >= N) continue;
        if(abs(pre - r) == 2) continue;//走回上一步了
        temp = state;
        state.MD -= MDT[tx * N + ty][state.f[tx * N + ty] - 1];//把tx*N+ty这个点的曼哈顿距离减去
        state.MD += MDT[sx * N + sy][state.f[tx * N + ty] - 1];//重新计算再加上
        swap(state.f[tx * N + ty],state.f[sx * N + sy]);
        state.space = tx * N + ty;
        if(dfs(depth+1,r)) {path[depth] = r;return true;}
        state = temp;
    }
    return false ;
}

string iterative_deepening(Puzzle in) {
    in.MD = getAllMd(in);
    for(limit = in.MD;limit <= LIMIT;limit++) {
        state = in;
        if(dfs(0,-100)) {
            string ans = "";
            for(int i = 0;i < limit;i++) ans += dir[path[i]];
            return ans;
        }
    }
    return "unsolvable";
}
int main() {
    for(int i = 0;i < N2;i++) 
        for(int j = 0;j < N2;j++) 
            MDT[i][j] = abs(i/N - j/N) + abs(i%N - j%N);
    Puzzle in;
    for(int i = 0;i < N2;i++) {
        cin>>in.f[i];
        if(in.f[i] == 0) {
            in.f[i] = N2;
            in.space = i;
        }
    }
    string ans = iterative_deepening(in);
    cout<<ans<<endl;
    cout<<ans.size()<<endl; 
    return 0;
}

A*:

#include<bits/stdc++.h>
#define N 4
#define N2 16
using namespace std;
const int dx[4] = {0,-1,0,1};
const int dy[4] = {1,0,-1,0};
const char dir[4] = {'r','u','l','d'};
int MDT[N2][N2];
struct Puzzle {
    int f[N2],space,MD;
    int cost;
    bool operator < (const Puzzle &p)const {
        for(int i = 0;i < N2;i++) {
            if(f[i] == p.f[i]) continue;
            return f[i] < p.f[i];
        }
        return false;
    }
};

struct State{
    Puzzle puzzle;
    int estimated;
    bool operator < (const State &s) const {
        return estimated > s.estimated;
    }
};

int getAllMd(Puzzle pz) {
    int sum = 0;
    for(int i = 0;i < N2;i++) {
        if(pz.f[i] == N2) continue;
        sum += MDT[i][pz.f[i] - 1];
    }
    return sum;
}

int astar(Puzzle s) {
    priority_queue<State> PQ;
    s.MD = getAllMd(s);
    s.cost = 0;
    map<Puzzle,bool> V;
    Puzzle u,v;
    State init;
    init.puzzle = s;
    init.estimated = getAllMd(s);
    PQ.push(init);
    while(!PQ.empty()) {
        State st = PQ.top();PQ.pop();
        u = st.puzzle;
        if(u.MD == 0) return u.cost;
        V[u] = true;
        int sx = u.space / N;
        int sy = u.space % N;
        for(int r = 0;r < 4;r++) {
            int tx = sx + dx[r];
            int ty = sy + dy[r];
            if(tx < 0 || ty < 0 || tx >= N || ty >= N) continue;
            v = u;
            v.MD -= MDT[tx * N + ty][v.f[tx * N + ty] - 1];
            v.MD += MDT[sx * N + sy][v.f[tx * N + ty] - 1];
            swap(v.f[sx * N + sy],v.f[tx * N + ty]);
            v.space = tx * N + ty;
            if(!V[v]) {
                v.cost++;
                State now;
                now.puzzle = v;
                now.estimated = v.cost + v.MD;
                PQ.push(now);
            }
        }
    } 
    return -1;                                                                
}

int main() {
    for(int i = 0;i < N2;i++) {
        for(int j = 0;j < N2;j++) {
            MDT[i][j] = abs(i/N - j/N) + abs(i%N - j%N);
        }
    }
    Puzzle in;
    for(int i = 0;i < N2;i++) {
        cin>>in.f[i];
        if(in.f[i] == 0) {
            in.f[i] = N2;
            in.space = i;
        }
    }
    cout<<astar(in)<<endl;
    return 0;
}