| Time Limit: 2 second(s) | Memory Limit: 32 MB |
Finally you found the city of Gold. As you are fond of gold, you start collecting them. But there are so much gold that you are getting tired collecting them.
So, you want to find the minimum effort to collect all the gold.
You can describe the city as a 2D grid, where your initial position is marked by an 'x'. An empty place will be denoted by a '.'. And the cells which contain gold will be denoted by 'g'. In each move you can go to all 8 adjacent places inside the city.
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case will start with a blank line and two integers, m and n (0 < m, n < 20) denoting the row and columns of the city respectively. Each of the next m lines will contain n characters describing the city. There will be exactly one 'x' in the city and at most 15 gold positions.
Output
For each case of input you have to print the case number and the minimum steps you have to take to collect all the gold and go back to 'x'.
Sample Input |
Output for Sample Input |
|
2 5 5 x.... g.... g.... ..... g.... 5 5 x.... g.... g.... ..... ..... |
Case 1: 8 Case 2: 4 |
思路：状态压缩dp；
一开始用状压搜索来写的wa；
这题是个状态压缩dp，dp[i][j],表示在状态i下到j结束的最小的花费，将起始点放入一起状压，因为起点也是终点，那么最后答案就是dp[(1<<cn)-1][cn-1];cn为起始的标号，也是点的个数
还有可以8个方向，那么两个点的最小距离就是x,y两个方向距离中大的那个；
1 #include<stdio.h> 2 #include<algorithm> 3 #include<stdlib.h> 4 #include<iostream> 5 #include<math.h> 6 #include<string.h> 7 #include<queue> 8 using namespace std; 9 typedef long long LL; 10 char ma[22][22]; 11 int __ma[22][22]; 12 int dp[1<<16][22]; 13 typedef struct node 14 { 15 int x; 16 int y; 17 } ss; 18 ss ans[100]; 19 int d(int n,int m,int nn,int mm) 20 { 21 int x = abs(n-nn); 22 int y = abs(m-mm); 23 return max(x,y); 24 } 25 int main(void) 26 { 27 int n,m; 28 int T,__ca=0; 29 scanf("%d",&T); 30 while(T--) 31 { 32 __ca++; 33 int i ,j; 34 //memset(flag,0,sizeof(flag)); 35 scanf("%d %d",&n,&m); 36 for(i = 0; i < n; i++) 37 { 38 scanf("%s",ma[i]); 39 } 40 int __n,__m; 41 int cn=0; 42 for(i = 0; i < n; i++) 43 { 44 for(j = 0; j < m; j++) 45 { 46 if(ma[i][j] == 'x') 47 { 48 __n=i; 49 __m=j; 50 } 51 if(ma[i][j]=='g') 52 { 53 ans[cn].x = i; 54 ans[cn].y = j; 55 __ma[i][j] = cn++; 56 } 57 } 58 } 59 int ask=0; 60 for(i = 0; i < (1<<16); i++) 61 { 62 for(j = 0; j < 16; j++) 63 { 64 dp[i][j] = 1e9; 65 } 66 } 67 ans[cn].x = __n; 68 ans[cn].y = __m; 69 __ma[__n][__m] = cn++; 70 dp[0|(1<<(cn-1))][cn-1] = 0; 71 if( cn != 1) 72 { 73 for( i = 0; i <(1<<cn); i++) 74 { 75 for(int x = 0; x < cn; x++) 76 { if(i & (1<<x)) 77 for(int y = 0; y < cn; y++) 78 { 79 int di=d(ans[x].x,ans[x].y,ans[y].x,ans[y].y); 80 dp[i|(1<<y)][y] = min(dp[i|(1<<y)][y],dp[i][x]+di); 81 } 82 } 83 } 84 ask=dp[(1<<cn)-1][cn-1]; 85 } 86 printf("Case %d: ",__ca); 87 printf("%d ",ask); 88 } 89 return 0; 90 }