poj2243 hdu1372 floyid warshall

来源：互联网发布：网上看房用什么软件编辑：程序博客网时间：2024/05/21 06:46

可以通过图的连通性进行求解。floyid算法也是比较好理解的。但和bfs比较它的空间复杂度较高。

    #include <iostream>    #include <string>    using namespace std;    const int chess = 9;    int w[chess][chess][chess][chess];    bool position(int x,int y)    {    if(x > 0 && x < chess && y >0 && y <chess)    return true;    return false;    }    void init()    {    memset(w,-1,sizeof(w));    for(int x = 1;x < chess;x++)    for(int y = 1;y < chess;y++)    {    if(position(x + 1,y + 2))    w[x][y][x + 1][y + 2] = 1;    if(position(x + 2,y + 1))    w[x][y][x + 2][y + 1] = 1;    if(position(x - 1,y + 2))    w[x][y][x - 1][y + 2] = 1;    if(position(x -2,y +1))    w[x][y][x -2][y +1] = 1;    if(position(x + 1,y - 2))    w[x][y][x + 1][y - 2] = 1;    if(position(x + 2,y -1))    w[x][y][x + 2][y - 1] = 1;    if(position(x - 1,y - 2))    w[x][y][x - 1][y - 2] = 1;    if(position(x -2,y -1))    w[x][y][x -2][y -1] = 1;    }//for int y    for(int kx = 1;kx < chess;kx++)    for(int ky = 1; ky < chess;ky++)    for(int ix = 1;ix < chess ;ix++)    for(int iy = 1;iy < chess;iy++)    for(int jx = 1;jx < chess;jx++)    for(int jy = 1;jy < chess;jy++)    {    if(w[ix][iy][jx][jy] > w[ix][iy][kx][ky] + w[kx][ky][jx][jy] && w[ix][iy][kx][ky] >0 && w[kx][ky][jx][jy] > 0)    w[ix][iy][jx][jy] = w[ix][iy][kx][ky] + w[kx][ky][jx][jy];    else if(-1==w[ix][iy][jx][jy] && w[ix][iy][kx][ky] >0 && w[kx][ky][jx][jy] > 0 )    w[ix][iy][jx][jy] = w[ix][iy][kx][ky] + w[kx][ky][jx][jy];    }    }    int main()    {    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    string s1,s2;    int res;    init();    while(cin>>s1>>s2)    {    int x,y,m,n;    x = s1[0] - 'a' + 1;    y = s1[1] - '0';    m = s2[0] - 'a' + 1;    n = s2[1] - '0';    if(x==m && y==n)    res = 0;    else    res = w[x][y][m][n];    cout<<"To get from "<<s1<<" to "<<s2<<" takes "<<res<<" knight moves."<<endl;    }    return 0;    }

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