hdu5067——Harry And Dig Machine(dfs+剪枝)
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Problem Description
As we all know, Harry Porter learns magic at Hogwarts School. However, learning magical knowledge alone is insufficient to become a great magician. Sometimes, Harry also has to gain knowledge from other certain subjects, such as language, mathematics, English, and even algorithm.
Dumbledore, the headmaster of Hogwarts, is planning to construct a new teaching building in his school. The area he selects can be considered as an n*m grid, some (but no more than ten) cells of which might contain stones. We should remove the stones there in order to save place for the teaching building. However, the stones might be useful, so we just move them to the top-left cell. Taking it into account that Harry learned how to operate dig machine in Lanxiang School several years ago, Dumbledore decides to let him do this job and wants it done as quickly as possible. Harry needs one unit time to move his dig machine from one cell to the adjacent one. Yet skilled as he is, it takes no time for him to move stones into or out of the dig machine, which is big enough to carry infinite stones. Given Harry and his dig machine at the top-left cell in the beginning, if he wants to optimize his work, what is the minimal time Harry needs to finish it?
Input
They are sever test cases, you should process to the end of file. For each test case, there are two integers n and m.
Output
For each test case, just output one line that contains an integer indicate the minimal time that Harry can finish his job.
Sample Input
3 3
0 0 0
0 100 0
0 0 0
2 2
1 1
1 1
Sample Output
4
4
数据规模很小,所以dfs暴力出所有的石堆,再稍微剪一下就行了
#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <math.h>#include <algorithm>#include <queue>#include <iomanip>#define INF 0x3f3f3f3f#define MAXN 105#define Mod 99999999using namespace std;int a[MAXN][MAXN];struct Node{ int x,y;};Node stone[MAXN*MAXN];int cnt,vis[MAXN*MAXN],ans;int getdis(Node a,Node b){ return abs(a.x-b.x)+abs(a.y-b.y);}void dfs(int step,int pre,int sum){ if(sum>ans) return; if(step==cnt) { sum=sum+getdis(stone[pre],stone[0]); if(sum<ans) ans=sum; return; } for(int i=1;i<cnt;++i) { if(!vis[i]) { vis[i]=1; dfs(step+1,i,sum+getdis(stone[i],stone[pre])); vis[i]=0; } }}int main(){ int n,m; stone[0].x=1; stone[0].y=1; while(~scanf("%d%d",&n,&m)) { memset(a,0,sizeof(a)); cnt=1; for(int i=1; i<=n; ++i) for(int j=1; j<=m; ++j) { scanf("%d",&a[i][j]); if(a[i][j]>0) { stone[cnt].x=i; stone[cnt].y=j; cnt++; } } ans=INF; dfs(1,0,0); printf("%d\n",ans); } return 0;}
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