zoukankan html css js c++ java

POJ 1606 jugs(又是水壶问题)

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #define sc(x) scanf("%d",&(x))
  6 #define pf(x) printf("%d
", x)
  7 #define CL(x, y) memset(x, y, sizeof(x))
  8 using namespace std;
  9 const int MAX = 105;
 10 int front, rear;
 11 int used[MAX][MAX];
 12 char str[6][10]= {"fill A", "fill B", "empty A", "empty B", "pour A B", "pour B A"};
 13 struct node
 14 {
 15     int a, b;
 16     int n;
 17     int step;
 18     node *pre;
 19 } Q[MAX*MAX];   //数组模拟队列    queue<Node> q;
 20 void back(node a);
 21 void BFS(int a, int b, int c);
 22 int main()
 23 {
 24     int a,b,c;
 25     while(cin >> a >> b >> c)
 26     {
 27         CL(used, 0);
 28         BFS(a, b, c);
 29         //pf(Q[rear].step);
 30         back(Q[rear]);
 31         cout << "success" << endl;
 32     }
 33 }
 34 void back(node a)
 35 {
 36     if(a.pre != NULL)
 37     {
 38         //printf("%s
", str[a.n]);
 39         back(*(a.pre));//递归，找出初始操作
 40         printf("%s
", str[a.n]);
 41     }
 42     return ;
 43 }
 44 void BFS(int a, int b, int c)
 45 {
 46     front = 0;
 47     rear = 0;
 48     Q[front].a = 0;
 49     Q[front].b = 0;
 50     Q[front].pre = NULL;
 51     Q[front].step = 0;
 52     used[0][0] = 1;
 53     rear++;
 54     while(front != rear)
 55     {
 56         node now = Q[front];  //{0, 0, NULL, 0}
 57         used[now.a][now.b] = 1;
 58         if(!used[a][now.b] && now.a!=a)//fill A   将 A 加满
 59         {
 60             Q[rear].n = 0;    //记录执行的操作
 61             Q[rear].step = Q[front].step + 1;     //步骤 +1
 62             Q[rear].a = a;
 63             Q[rear].b = now.b;
 64             Q[rear].pre = &Q[front];
 65             used[a][now.b] = 1;
 66             if(a==c || now.b==c) break; //完成操作
 67             rear++;
 68         }
 69         if(!used[now.a][b] && now.b!=b)//fill B   将 B 加满
 70         {
 71             Q[rear].n = 1;
 72             Q[rear].step = Q[front].step+1;
 73             Q[rear].a = now.a;
 74             Q[rear].b = b;
 75             Q[rear].pre = &Q[front];
 76             used[now.a][b] = 1;
 77             if(now.a==c || b==c) break;
 78             rear++;
 79         }
 80         if(!used[0][now.b] && now.a!=0)//empty A
 81         {
 82             Q[rear].n = 2;
 83             Q[rear].step = Q[front].step + 1;
 84             Q[rear].a = 0;
 85             Q[rear].b = now.b;
 86             Q[rear].pre = &Q[front];
 87             used[0][now.b] = 1;
 88             if(now.b == c) break;//A中没有水
 89             rear++;
 90         }
 91         if(!used[now.a][0] && now.b!=0)//empty B
 92         {
 93             Q[rear].n = 3;
 94             Q[rear].step = Q[front].step + 1;
 95             Q[rear].a = now.a;
 96             Q[rear].b = 0;
 97             Q[rear].pre = &Q[front];
 98             used[now.a][0] = 1;
 99             if(now.a == c) break;//B中没有水
100             rear++;
101         }
102         int x = now.a < b-now.b ? now.a : b-now.b;
103         if(!used[now.a-x][now.b+x])//pour A B
104         {
105             Q[rear].n = 4;
106             Q[rear].step = Q[front].step+1;
107             Q[rear].a = now.a - x;
108             Q[rear].b = now.b + x;
109             Q[rear].pre = &Q[front];
110             used[now.a-x][now.b+x] = 1;
111             if(now.a-x==c || now.b+x==c) break;
112             rear++;
113         }
114         int y = a-now.a < now.b ? a-now.a : now.b;//找到与自己本壶最接近的到水量
115         if(!used[now.a+y][now.b-y])//pour B A
116         {
117             Q[rear].n = 5;
118             Q[rear].step = Q[front].step+1;
119             Q[rear].a = now.a+y;
120             Q[rear].b = now.b-y;
121             Q[rear].pre = &Q[front];
122             used[now.a+y][now.b-y] = 1;
123             if(now.a+y==c || now.b-y==c) break;
124             rear++;
125         }
126         front++;
127     }//跳出循环时， front==rear
128     //if(front == rear) printf("impossible
");//条件判断
129 }

View Code

上面解法不完整，但是还是在POJ上面过了

应该是使得B中存留那么多水

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #define sc(x) scanf("%d",&(x))
  6 #define pf(x) printf("%d
", x)
  7 #define CL(x, y) memset(x, y, sizeof(x))
  8 using namespace std;
  9 const int MAX = 105;
 10 int front, rear;
 11 int used[MAX][MAX];
 12 char str[6][10]= {"fill A", "fill B", "empty A", "empty B", "pour A B", "pour B A"};
 13 struct node
 14 {
 15     int a, b;
 16     int n;
 17     int step;
 18     node *pre;
 19 } Q[MAX*MAX];   //数组模拟队列    queue<Node> q;
 20 void back(node a);
 21 void BFS(int a, int b, int c);
 22 int main()
 23 {
 24     int a,b,c;
 25     while(cin >> a >> b >> c)
 26     {
 27         CL(used, 0);
 28         BFS(a, b, c);
 29         //pf(Q[rear].step);
 30         back(Q[rear]);
 31         cout << "success" << endl;
 32     }
 33 }
 34 void back(node a)
 35 {
 36     if(a.pre != NULL)
 37     {
 38         //printf("%s
", str[a.n]);
 39         back(*(a.pre));//递归，找出初始操作
 40         printf("%s
", str[a.n]);
 41     }
 42     return ;
 43 }
 44 void BFS(int a, int b, int c)
 45 {
 46     front = 0;
 47     rear = 0;
 48     Q[front].a = 0;
 49     Q[front].b = 0;
 50     Q[front].pre = NULL;
 51     Q[front].step = 0;
 52     used[0][0] = 1;
 53     rear++;
 54     while(front != rear)
 55     {
 56         node now = Q[front];  //{0, 0, NULL, 0}
 57         used[now.a][now.b] = 1;
 58         if(!used[a][now.b] && now.a!=a)//fill A   将 A 加满
 59         {
 60             Q[rear].n = 0;    //记录执行的操作
 61             Q[rear].step = Q[front].step + 1;     //步骤 +1
 62             Q[rear].a = a;
 63             Q[rear].b = now.b;
 64             Q[rear].pre = &Q[front];
 65             used[a][now.b] = 1;
 66             //if(a==c || now.b==c) break; //完成操作
 67             if(now.b==c) break; //完成操作
 68             rear++;
 69         }
 70         if(!used[now.a][b] && now.b!=b)//fill B   将 B 加满
 71         {
 72             Q[rear].n = 1;
 73             Q[rear].step = Q[front].step+1;
 74             Q[rear].a = now.a;
 75             Q[rear].b = b;
 76             Q[rear].pre = &Q[front];
 77             used[now.a][b] = 1;
 78             //if(now.a==c || b==c) break;
 79             if(b==c) break;
 80             rear++;
 81         }
 82         if(!used[0][now.b] && now.a!=0)//empty A
 83         {
 84             Q[rear].n = 2;
 85             Q[rear].step = Q[front].step + 1;
 86             Q[rear].a = 0;
 87             Q[rear].b = now.b;
 88             Q[rear].pre = &Q[front];
 89             used[0][now.b] = 1;
 90             if(now.b == c) break;//A中没有水
 91             rear++;
 92         }
 93         if(!used[now.a][0] && now.b!=0)//empty B
 94         {
 95             Q[rear].n = 3;
 96             Q[rear].step = Q[front].step + 1;
 97             Q[rear].a = now.a;
 98             Q[rear].b = 0;
 99             Q[rear].pre = &Q[front];
100             used[now.a][0] = 1;
101           //  if(now.a == c) break;//B中没有水
102             rear++;
103         }
104         int x = now.a < b-now.b ? now.a : b-now.b;
105         if(!used[now.a-x][now.b+x])//pour A B
106         {
107             Q[rear].n = 4;
108             Q[rear].step = Q[front].step+1;
109             Q[rear].a = now.a - x;
110             Q[rear].b = now.b + x;
111             Q[rear].pre = &Q[front];
112             used[now.a-x][now.b+x] = 1;
113             //if(now.a-x==c || now.b+x==c) break;
114             if(now.b+x==c) break;
115             rear++;
116         }
117         int y = a-now.a < now.b ? a-now.a : now.b;//找到与自己本壶最接近的到水量
118         if(!used[now.a+y][now.b-y])//pour B A
119         {
120             Q[rear].n = 5;
121             Q[rear].step = Q[front].step+1;
122             Q[rear].a = now.a+y;
123             Q[rear].b = now.b-y;
124             Q[rear].pre = &Q[front];
125             used[now.a+y][now.b-y] = 1;
126             //if(now.a+y==c || now.b-y==c) break;
127             if(now.b-y==c) break;
128             rear++;
129         }
130         front++;
131     }//跳出循环时， front==rear
132     //if(front == rear) printf("impossible
");//条件判断
133 }

View Code

查看全文

相关阅读:
POJ 1795 DNA Laboratory
CodeForces 303B Rectangle Puzzle II
HDU 2197 本源串
 HDU 5965 扫雷
 POJ 3099 Go Go Gorelians
CodeForces 762D Maximum path
CodeForces 731C Socks
HDU 1231 最大连续子序列
 HDU 5650 so easy
大话接口隐私与安全转载

原文地址：https://www.cnblogs.com/ghostTao/p/4305596.html