不知道為什么用二維數(shù)組來實(shí)現(xiàn)不能，留給未來的我來解決吧。

#define N 3 #define M 4#include <iostream> #include<string> #include<queue> #include<map> using namespace std; struct Puzzle {int grids[N][N];//拼圖坐標(biāo)的二維數(shù)組描述string path = "";//路徑int space_x;int space_y;//目標(biāo)點(diǎn)的坐標(biāo)bool operator < (const Puzzle &p) const{for (int i = 0; i < N; i++){for (int j = 0; j < N; j++){if (grids[i][j] == p.grids[i][j]) continue;return grids[i][j]< p.grids[i][j];}}return false;} }; //確立方向的變化 static const char dir[M] = {'L','R','U','D'}; static const int dx[M] = {0, 0, -1, 1}; static const int dy[M] = {-1, 1, 0, 0}; bool is_target(Puzzle tp)//判斷是否到達(dá)目標(biāo)狀態(tài) {for (int i = 0; i < N; i++){for (int j = 0; j < N; j++){if (tp.grids[i][j] != (3 * i + j + 1)){return false;}}}return true; }string bfs(Puzzle tp)//用寬度搜索進(jìn)行狀態(tài)轉(zhuǎn)移 {Puzzle tp2;queue<Puzzle> qu;map<Puzzle, bool>Cut;tp.path = "";qu.push(tp);Cut[tp] = true;while(!qu.empty()){printf("1\n");tp2 = qu.front(); qu.pop();if (is_target(tp2)) {return tp2.path;}for (int i = 0; i < M; i++){int space_x = tp2.space_x+dx[i];int space_y = tp2.space_y+dy[i];if (space_x < 0 || space_y < 0 || space_x >= N || space_y >= N){continue;}swap(tp2.grids[tp2.space_x][tp2.space_y], tp2.grids[space_x][space_y]);tp2.space_x = space_x;tp2.space_y = space_y;if (!Cut[tp2]){Cut[tp2] = true;tp2.path += dir[i];qu.push(tp2);}}}return "unsolvable"; }int main() {Puzzle sp;//初始化拼圖坐標(biāo)for (int i = 0; i < N; i++){for (int j = 0; j < N; j++){cin>>sp.grids[i][j];if (sp.grids[i][j] == 0){sp.grids[i][j] = N*3;sp.space_x = i;sp.space_y = j;}}}for (int i = 0; i < N; i++){for (int j = 0; j < N; j++){cout << sp.grids[i][j]<<" ";}cout<<"\n";}cout << is_target(sp)<<"\n";string ans = bfs(sp);cout << ans;system("pause");return 0; }

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