[Sicily 1153 马的周游问题] 优化思想：改变搜索顺序

来源：互联网发布：java不换行编辑：程序博客网时间：2024/05/01 10:14
#include<iostream> #include<vector>//1153马的周游问题：（优化基本思想）先搜索可行格较少的格子 #include<algorithm> using namespace std;const int SIZE = 65;int visited[SIZE];int route[SIZE];int cnt_size[SIZE];int dx[8] = {-1, -2, -2, -1, 1, 2, 2, 1};int dy[8] = {-2, -1, 1, 2, -2, -1, 1, 2};vector<int> neighbors[SIZE];//neighbors[i][j]记录的是马可以从i到j bool cmp(int x, int y){return cnt_size[x] < cnt_size[y];}int get_child_size(int child){int size = 0;for(int i = 0; i < neighbors[child].size(); i++){if(!visited[neighbors[child][i]])    size++;}return size;}void get_children(int x, vector<int> &children){    for(int i = 0; i < neighbors[x].size(); i++)    {        int child = neighbors[x][i];        if(!visited[child])        {            children.push_back(child);            cnt_size[child] = get_child_size(child);        }    }    sort(children.begin(), children.end(), cmp);//排序，可行格较少的排在前面 }void print_route(){    for(int i = 1; i < SIZE - 1; i++)        cout << route[i] << " ";    cout << route[SIZE - 1] << endl;}bool dfs(int x, int depth){    route[depth] = x;//route[]记录路径     if(depth == (SIZE - 1))    {        print_route();        return true;    }    visited[x] = 1;    vector<int> children;//存放x点的还未被visit的neighbors     get_children(x, children);    for(int i = 0; i < children.size(); i++)    {        if(dfs(children[i], depth + 1))//如果某一个点可以继续搜下去，就返回true             return true;        //如果不能继续搜下去，就搜索下一个children     }    visited[x] = false;//如果这个点的所有children都搜索不下去，则回溯     return false;}int main(){    int start;//马的起点为start，范围是1到64     while(cin >> start && start != -1)    {    //初始化         for(int i = 0; i < SIZE; i++)            visited[i] = 0;                for(int i = 0; i < SIZE; i++)            neighbors[i].clear();        for(int i = 1; i < SIZE; i++)        {            int row = (i + 7) / 8;//计算i点在第几行             int column = i - (row - 1)* 8;//计算i点在第几列             for(int j = 0; j < 8; j++)//对于每一个i点，计算出它的neighbors             {                int nx = row + dx[j];                int ny = column + dy[j];                if(nx >= 1 && nx <= 8 && ny >=1 && ny <= 8)                    neighbors[i].push_back((nx - 1) * 8 + ny);            }        }        dfs(start, 1);    }    return 0;}
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