题意:一个简单图,给你一个起点,图中最多有4个宝藏,问你收集这四个宝藏最多要多少步.
解题思路:很简单的一个题,因为宝藏比较少,所以先对宝藏进行一个全排,然后再一步一步算步数.
解题代码:
1 // File Name: d.cpp 2 // Author: darkdream 3 // Created Time: 2014年05月06日 星期二 19时03分57秒 4 5 #include<vector> 6 #include<set> 7 #include<deque> 8 #include<stack> 9 #include<bitset> 10 #include<algorithm> 11 #include<functional> 12 #include<numeric> 13 #include<utility> 14 #include<sstream> 15 #include<iostream> 16 #include<iomanip> 17 #include<cstdio> 18 #include<cmath> 19 #include<cstdlib> 20 #include<cstring> 21 #include<ctime> 22 23 using namespace std; 24 int n , m; 25 struct node{ 26 int x, y ,s; 27 }LIST[10005]; 28 int visit[104][104]; 29 int MAP[104][104]; 30 struct node1{ 31 int x, y ; 32 }a[10]; 33 int xadd[] = {0,0,1,-1}; 34 int yadd[] = {1,-1,0,0}; 35 int ok = 1; 36 int bfs(int bx, int by ,int ex,int ey) 37 { 38 //printf("%d %d %d %d ",bx,by,ex,ey); 39 LIST[1].x = bx ; 40 LIST[1].y = by ; 41 LIST[1].s = 0 ; 42 int l = 1, r= 1; 43 int tempok = 0 ; 44 memset(visit,0,sizeof(visit)); 45 visit[bx][by] = 1; 46 while(l <= r ) 47 { 48 // printf("%d %d ",LIST[l].x,LIST[l].y); 49 if(LIST[l].x == ex && LIST[l].y == ey) 50 { 51 tempok = 1; 52 break; 53 } 54 for(int i = 0 ; i <= 3;i ++) 55 { 56 int tempx = LIST[l].x + xadd[i]; 57 int tempy = LIST[l].y + yadd[i]; 58 if(!visit[tempx][tempy] && MAP[tempx][tempy]) 59 { 60 visit[tempx][tempy] = 1; 61 r ++ ; 62 LIST[r].x = tempx; 63 LIST[r].y = tempy; 64 LIST[r].s = LIST[l].s +1 ; 65 } 66 } 67 l ++; 68 } 69 if(tempok == 0) 70 ok = 0; 71 else return LIST[l].s; 72 73 } 74 int main(){ 75 76 77 while(scanf("%d %d",&n,&m) != EOF) 78 { 79 if(n == 0 && m == 0) 80 break; 81 memset(MAP,0,sizeof(MAP)); 82 memset(a,0,sizeof(a)); 83 for(int i = 1;i <= n;i ++) 84 { 85 char str[104]; 86 scanf("%s",&str[1]); 87 for(int j = 1;j <= m;j ++) 88 { 89 if(str[j] == '#') 90 { 91 } 92 else 93 { 94 MAP[i][j] = 1; 95 if(str[j] == '@') 96 { 97 a[0].x = i ; 98 a[0].y = j; 99 } 100 } 101 } 102 } 103 int k ; 104 scanf("%d",&k); 105 for(int i = 1;i <= k;i ++) 106 scanf("%d %d",&a[i].x,&a[i].y); 107 int temp[6]; 108 for(int i =0 ;i < k;i ++) 109 temp[i] = i + 1; 110 ok = 1; 111 int min = 1000000; 112 do{ 113 int last = 0; 114 int sum = 0 ; 115 for(int i = 0 ;i < k ;i ++) 116 { 117 sum += bfs(a[last].x,a[last].y,a[temp[i]].x,a[temp[i]].y); 118 last = temp[i]; 119 } 120 if (sum <= min) 121 min = sum; 122 123 // printf("%d ",sum); 124 }while(next_permutation(temp,temp+k)); 125 if(ok == 0 ) 126 printf("-1 "); 127 else 128 printf("%d ",min); 129 } 130 131 return 0 ; 132 }