HDU 1198 Farm Irrigation

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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1198

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6023    Accepted Submission(s): 2602


Problem Description
Benny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. Water pipes are placed in these squares. Different square has a different type of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows.


Figure 1


Benny has a map of his farm, which is an array of marks denoting the distribution of water pipes over the whole farm. For example, if he has a map

ADC
FJK
IHE

then the water pipes are distributed like


Figure 2


Several wellsprings are found in the center of some squares, so water can flow along the pipes from one square to another. If water flow crosses one square, the whole farm land in this square is irrigated and will have a good harvest in autumn.

Now Benny wants to know at least how many wellsprings should be found to have the whole farm land irrigated. Can you help him?

Note: In the above example, at least 3 wellsprings are needed, as those red points in Figure 2 show.
 

Input
There are several test cases! In each test case, the first line contains 2 integers M and N, then M lines follow. In each of these lines, there are N characters, in the range of 'A' to 'K', denoting the type of water pipe over the corresponding square. A negative M or N denotes the end of input, else you can assume 1 <= M, N <= 50.
 

Output
For each test case, output in one line the least number of wellsprings needed.
 

Sample Input
2 2DKHF3 3ADCFJKIHE-1 -1
 

Sample Output
23
 深搜,看上去好像很复杂的样子,其实除了建图,处理方式就跟常规题一样,建图只需要考虑相邻直接能不能到达,可以用一个三维数组存储下来。
建图问题解决了,那就是普通的深搜,所以,这题也是道水题~
#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<sstream>#include<vector>#include<map>#include<list>#include<set>#include<queue>#define LL long longusing namespace std;const int maxn=1005,maxe=100005,inf=1<<29;int dir[][2]={{0,1},{-1,0},{0,-1},{1,0}};int mat[maxn][maxn],n,m;bool vis[maxn][maxn];int Map[11][11][4]={    {{0,0,0,0},{0,0,1,0},{0,1,0,0},{0,1,1,0},{0,1,0,0},{0,0,1,0},{0,0,1,0},{0,1,0,0},{0,1,1,0},{0,1,1,0},{0,1,1,0}},    {{1,0,0,0},{0,0,0,0},{1,1,0,0},{0,1,0,0},{0,1,0,0},{1,0,0,0},{1,0,0,0},{1,1,0,0},{1,1,0,0},{0,1,0,0},{1,1,0,0}},    {{0,0,0,1},{0,0,1,1},{0,0,0,0},{0,0,1,0},{0,0,0,1},{0,0,1,0},{0,0,1,1},{0,0,0,1},{0,0,1,0},{0,0,1,1},{0,0,1,1}},    {{1,0,0,1},{0,0,0,1},{1,0,0,0},{0,0,0,0},{0,0,0,1},{1,0,0,0},{1,0,0,1},{1,0,0,1},{1,0,0,0},{0,0,0,1},{1,0,0,1}},    {{0,0,0,1},{0,0,0,1},{0,1,0,0},{0,1,0,0},{0,1,0,1},{0,0,0,0},{0,0,0,1},{0,1,0,1},{0,1,0,0},{0,1,0,1},{0,1,0,1}},    {{1,0,0,0},{0,0,1,0},{1,0,0,0},{0,0,1,0},{0,0,0,0},{1,0,1,0},{1,0,1,0},{1,0,0,0},{1,0,1,0},{0,0,1,0},{1,0,1,0}},    {{1,0,0,0},{0,0,1,0},{1,1,0,0},{0,1,1,0},{0,1,0,0},{1,0,1,0},{1,0,1,0},{1,1,0,0},{1,1,1,0},{0,1,1,0},{1,1,1,0}},    {{0,0,0,1},{0,0,1,1},{0,1,0,0},{0,1,1,0},{0,1,0,1},{0,0,1,0},{0,0,1,1},{0,1,0,1},{0,1,1,0},{0,1,1,1},{0,1,1,1}},    {{1,0,0,1},{0,0,1,1},{1,0,0,0},{0,0,1,0},{0,0,0,1},{1,0,1,0},{1,0,1,1},{1,0,0,1},{1,0,1,0},{0,0,1,1},{1,0,1,1}},    {{1,0,0,1},{0,0,0,1},{1,1,0,0},{0,1,0,0},{0,1,0,1},{1,0,0,0},{1,0,0,1},{1,1,0,1},{1,1,0,0},{0,1,0,1},{1,1,0,1}},    {{1,0,0,1},{0,0,1,1},{1,1,0,0},{0,1,1,0},{0,1,0,1},{1,0,1,0},{1,0,1,1},{1,1,0,1},{1,1,1,0},{0,1,1,1},{1,1,1,1}}};void dfs(int x,int y){    vis[x][y]=1;    for(int i=0;i<4;i++)    {        int nx=dir[i][0]+x,ny=dir[i][1]+y;        if(0<=nx&&nx<n&&0<=ny&&ny<m&&!vis[nx][ny]           &&Map[mat[x][y]][mat[nx][ny]][i]) dfs(nx,ny);    }}int main(){    while(~scanf("%d%d",&n,&m)&&!(n==-1&&m==-1))    {        char str[100];        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            scanf("%s",str);            for(int j=0;j<m;j++)                mat[i][j]=str[j]-'A';        }        int ans=0;        for(int i=0;i<n;i++)            for(int j=0;j<m;j++)            {                if(!vis[i][j])                {                    ans++;                    dfs(i,j);                }            }        printf("%d\n",ans);    }    return 0;}



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