ural 1400. Cellular Characters

题意：有n个字符，第一列有m个，类推下去。选择一个光标位置，使得他到每一个字符的期望最小。

第一行不能向上走，最下行不能向下走，其他跟诺基亚的类似。求最小期望。并输出最短路。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#include <stack>
#include <map>

using namespace std;
const int N = 1009;
int n,m;
struct nod{
    int x,y;
}nex[N][N][4];
int dis[N][N];
const int INF = 0x3f3f3f3f;
void init()
{

    int row = (n-1)/m+1,las = (n-1)%m+1;
    for(int i=0;i<row;i++)
    for(int j=0;j<m;j++)
    {
        nex[i][j][0].x = i,nex[i][j][0].y = j-1;
        nex[i][j][1].x = i+1,nex[i][j][1].y = j;
        nex[i][j][2].x = i,nex[i][j][2].y = j+1;
        nex[i][j][3].x = i-1,nex[i][j][3].y = j;
    }
    //0
    nex[0][0][0].x = row-1,nex[0][0][0].y = las-1;
    for(int i=1;i<row;i++) nex[i][0][0].x = i-1,nex[i][0][0].y = m-1;
    //2
    for(int i=0;i<row;i++) nex[i][m-1][2].x = i+1,nex[i][m-1][2].y = 0;
    //1,3
    if(row==1)
    {
        for(int i=0;i<n;i++)
        nex[0][i][1].x =0,nex[0][i][3].x = 0;
    }
    else
    {
        //1
        for(int i=0;i<m;i++)
        if(i<las)
        {
            nex[row-1][i][1].x = row-1,nex[row-1][i][1].y = i;
        }else
        {
            nex[row-2][i][1].x = row-2,nex[row-2][i][1].y = i;
        }
        //3
        for(int i=0;i<m;i++)
        if(i<las)
        {
            nex[0][i][3].x = 0,nex[0][i][3].y = i;
        }else
        {
            nex[0][i][3].x = 0,nex[0][i][3].y = i;
        }
    }
    //last
    nex[row-1][las][2].x = 0,nex[row-1][las][2].y = 0;
}
queue<nod> que;
nod ans;
int mindis(int x,int y)
{
    while(!que.empty()) que.pop();
    nod e,t;
    e.x = x,e.y = y;
    que.push(e);
    memset(dis,INF,sizeof(dis));
    dis[e.x][e.y] = 0;
    while(!que.empty())
    {
        e = que.front();
        que.pop();
        int d = dis[e.x][e.y];
        int tx,ty;
        for(int i=0;i<4;i++)
        {
            tx = nex[e.x][e.y][i].x;
            ty = nex[e.x][e.y][i].y;
            if(dis[tx][ty]>d+1)
            {
                dis[tx][ty] =d+1;
                t.x = tx,t.y = ty;
                que.push(t);
            }
        }
    }
    int ret = 0;
    for(int i=0;i<n;i++)
    {
        ret+=dis[i/m][i%m];
    }
    return ret;
}
void solve()
{
    int tmp = INF;
    for(int i=0;i<n;i++)
    {
        int x = i/m,y = i%m;
        memset(dis,INF,sizeof(dis));
        int k =mindis(x,y);
        if(tmp>k)
        {
            tmp = k;
            ans.x = x,ans.y =y;
        }
    }
    printf("Mean = %.2lf\n",1.0*tmp/n);
    mindis(ans.x,ans.y);
    for(int i=0;i<n;i++)
    {
        if(i&&i%m==0) printf("\n");
        if(i%m) printf(" ");
        printf("%d",dis[i/m][i%m]);
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    scanf("%d%d",&n,&m);
    if(m>n) m = n;
    init();
    solve();
    return 0;
}