HDU 2612

#include<iostream>#include<cstdio>#include<algorithm>#include<queue>#include<cstring>using namespace std;#define N 205#define inf 0x3f3f3f3fchar s[N][N];int vis[N][N];int sx,sy,ex,ey;int vx[]={0,0,1,-1};int vy[]={1,-1,0,0};int ans1[N][N],ans2[N][N];int n,m;struct point {int x,y,step;point (int xx,int yy,int ss){x=xx,y=yy,step=ss;}};void bfs(int judge){memset(vis,0,sizeof(vis));memset(ans1,inf,sizeof(ans1));memset(ans2,inf,sizeof(ans2));queue<point>q;q.push(point(sx,sy,0));while(!q.empty()){int ok=0;point top=q.front();q.pop();if(s[top.x][top.y]=='@'){ans1[top.x][top.y]=top.step;++ok;if(ok==judge)break;}for(int i=0;i<4;i++){int tx=top.x+vx[i];int ty=top.y+vy[i];if(tx<0||ty<0||tx>=n||ty>=m||vis[tx][ty]||s[tx][ty]=='#')continue;q.push(point(tx,ty,top.step+1));vis[tx][ty]=1;}}while(!q.empty())q.pop();memset(vis,0,sizeof(vis));q.push(point(ex,ey,0));while(!q.empty()){int ok=0;point top=q.front();q.pop();if(s[top.x][top.y]=='@'){ans2[top.x][top.y]=top.step;++ok;if(ok==judge)break;}for(int i=0;i<4;i++){int tx=top.x+vx[i];int ty=top.y+vy[i];if(tx<0||ty<0||tx>=n||ty>=m||vis[tx][ty]||s[tx][ty]=='#')continue;q.push(point(tx,ty,top.step+1));vis[tx][ty]=1;}}}int main(){#ifdef CDZSCfreopen("i.txt","r",stdin);#endifwhile(~scanf("%d%d",&n,&m)){if(n==0&&m==0)break;int ans=inf;for(int i=0;i<n;i++){for(int j=0;j<m;j++){cin>>s[i][j];if(s[i][j]=='Y'){sx=i,sy=j;}if(s[i][j]=='M'){ex=i,ey=j;}}}int sum=0;for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(s[i][j]=='@'){sum++;}}}bfs(sum);for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(s[i][j]=='@'){ans=min(ans,ans1[i][j]+ans2[i][j]);}}}printf("%d\n",ans*11);}return 0;}

题意给你2个起点‘M’和‘Y’，求这两个点到‘@’这个字符总和距离最小，‘@’会有多个。先将‘@’的个数计数为sum。这样一看好像要调用sum次BFS，其实仔细分析可以知道每次调用BFS所有的最短路已经被出来了，不必多次调用了。那么我们可以用二维数组保存当前的最短路，如果遇到‘@’这个点，下标为x，y。就保存ans[]x[y]=step;这样只要进行2次BFS就可以求出'M'和'Y'到所有‘@’的最短路了

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7029    Accepted Submission(s): 2334

Problem Description

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. 
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.

 

Input

The input contains multiple test cases.
Each test case include, first two integers n, m. (2<=n,m<=200). 
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’    express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KCF

 

Output

For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.

 

Sample Input

4 4Y.#@.....#..@..M4 4Y.#@.....#..@#.M5 5Y..@..#....#...@..M.#...#

 

Sample Output

668866

 

Author

yifenfei

 

Source

奋斗的年代

 

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