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;}