hdu 2579 Dating with girls(2)

来源：互联网发布：湖北农村金融数据编辑：程序博客网时间：2024/05/22 03:51

k<=10 用三维数组，第三维标记对k的余数，这样能把每种状态标记到。

#include<stdio.h>#include<string.h>char a[120][120];struct f{    int x,y,step;}b[120*120*4];int book[120][120][10];int main(){ int t; scanf("%d",&t); while(t--){    int n,m,k;    int startx=0,starty=0,endx=0,endy=0;    scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=n;i++)  scanf("%s",a[i]+1);for(int i=1;i<=n;i++)   for(int j=1;j<=m;j++)    {        if(a[i][j]=='Y'){startx=i,starty=j;}        if(a[i][j]=='G'){endx=i,endy=j;}    }    if((startx==0&&starty==0)||(endx==0&&endy==0)){printf("Please give me another chance!\n");continue;}    int head=0,tail=1,flag=0;    memset(book,0,sizeof(book));    b[head].x=startx,b[head].y=starty,b[head].step=0;    book[startx][starty][1]=1;    while(head<tail){            int yy=b[head].step+1;            int next[4][2]={{0,1},{1,0},{0,-1},{-1,0}};        for(int i=0;i<4;i++)        {            int tx=b[head].x+next[i][0],ty=b[head].y+next[i][1];            if(tx<1||tx>n||ty<1||ty>m||book[tx][ty][yy%k]) continue;            if(a[tx][ty]!='#'||yy%k==0){             b[tail].x=tx,b[tail].y=ty,b[tail++].step=yy;             book[tx][ty][yy%k]=1;              if(tx==endx&&ty==endy) {flag=1;break;}            }        }        if(flag) break;        head++;    }    if(flag){printf("%d\n",b[tail-1].step);}    else printf("Please give me another chance!\n"); }return 0;}

开始，3维数组的第三维大小开的是2，0代表障碍物全部消失，1代表全部存在，但有数据过不了(标记的状态少了)；

#include<stdio.h>#include<string.h>char a[120][120];struct f{    int x,y,step;}b[120*120*4];int book[120][120][2];int main(){ int t; scanf("%d",&t); while(t--){    int n,m,k;    int startx=0,starty=0,endx=0,endy=0;    scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=n;i++)  scanf("%s",a[i]+1);for(int i=1;i<=n;i++)   for(int j=1;j<=m;j++)    {        if(a[i][j]=='Y'){startx=i,starty=j;}        if(a[i][j]=='G'){endx=i,endy=j;}    }    if(startx==0&&starty==0||endx==0&&endy==0){printf("Please give me another chance!\n");continue;}    int head=0,tail=1,flag=0;    memset(book,0,sizeof(book));    b[head].x=startx,b[head].y=starty,b[head].step=0;    book[startx][starty][1]=1;    while(head<tail){            int yy=b[head].step+1;            int next[4][2]={{0,1},{1,0},{0,-1},{-1,0}};        for(int i=0;i<4;i++)        {            int tx=b[head].x+next[i][0],ty=b[head].y+next[i][1];            if(tx<1||tx>n||ty<1||ty>m) continue;            if(yy%k==0&&book[tx][ty][0]) continue;            if(yy%k!=0&&book[tx][ty][1]) continue;            if(a[tx][ty]!='#'||yy%k==0){                if(yy%k==0){book[tx][ty][0]=1;}                else {book[tx][ty][1]=1;}            b[tail].x=tx,b[tail].y=ty,b[tail++].step=yy;              if(tx==endx&&ty==endy) {flag=1;break;}            }        }        if(flag) break;        head++;    }    if(flag){printf("%d\n",b[tail-1].step);}    else printf("Please give me another chance!\n"); }return 0;}

12 4 4Y.#G#.##

0 0