hdu 1509 & hdu 1873 & hdu 1896 (基础优先队列)

http://acm.hdu.edu.cn/showproblem.php?pid=1509

裸的优先队列的应用,输入PUT的时候输入名字,值和优先值进队列,输入GRT的时候输出优先值小的名字和对应的值

注意的是优先级一样的时候输出顺序在前的

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 using namespace std;
 5 struct point {
 6     int val,odr,num;
 7     char na[101];
 8     bool operator <(const point & q)const
 9     {
10         if (odr==q.odr) return num>q.num;
11         else return odr>q.odr;
12     }
13 };
14 int main()
15 {
16     char lsy[5];
17     int k=0;
18     point temp;
19     priority_queue<point>que;
20     while (~scanf("%s",lsy))
21     {
22         if (strcmp(lsy,"GET")==0)
23         {
24             if (que.size()!=0)
25             {
26                 temp=que.top();
27                 que.pop();
28                 printf("%s %d\n",temp.na,temp.val);
29             }
30             else
31                 printf("EMPTY QUEUE!\n");
32         }
33         else
34         {
35             scanf("%s %d %d",temp.na,&temp.val,&temp.odr);
36             temp.num=k++;
37             que.push(temp);
38         }
39     }
40     return 0;
41 }

http://acm.hdu.edu.cn/showproblem.php?pid=1873

也很简单的优先队列,有三个医生的,每个医生又自己的顺序,所以可以建立三个队列

 1 #include<cstdio>
 2 #include<queue>
 3 #include<cstring>
 4 using namespace std;
 5 struct point {
 6     int x,num;
 7     bool operator <(const point & q)const
 8     {
 9         if (x==q.x) return num>q.num;
10         else return x<q.x;
11     }
12 };
13 int main()
14 {
15     int t,x,y;
16     char lsy[5];
17     priority_queue<point> a,b,c;
18     point temp;
19     while (~scanf("%d",&t))
20     {
21         int k=1;
22         while (a.size()!=0)
23            a.pop();
24         while (b.size()!=0)
25            b.pop();
26         while (c.size()!=0)
27            c.pop();
28         while (t--)
29         {
30             scanf("%s",lsy);
31             if (strcmp(lsy,"IN")==0)
32             {
33                 scanf("%d %d",&x,&y);
34                 temp.num=k++;
35                 temp.x=y;
36                 if (x==1) a.push(temp);
37                 else if (x==2) b.push(temp);
38                 else c.push(temp);
39             }
40             else
41             {
42                 scanf("%d",&x);
43                 if (x==1)
44                 {
45                     if (a.size()==0)
46                         printf("EMPTY\n");
47                     else
48                     {
49                         temp=a.top(),a.pop();
50                         printf("%d\n",temp.num);
51                     }
52                 }
53                 else if (x==2)
54                 {
55                     if (b.size()==0)
56                         printf("EMPTY\n");
57                     else
58                     {
59                         temp=b.top(),b.pop();
60                         printf("%d\n",temp.num);
61                     }
62                 }
63                 else
64                 {
65                     if (c.size()==0)
66                         printf("EMPTY\n");
67                     else
68                     {
69                         temp=c.top(),c.pop();
70                         printf("%d\n",temp.num);
71                     }
72                 }
73             }
74         }
75     }
76     return 0;
77 }

http://acm.hdu.edu.cn/showproblem.php?pid=1896

题意 遇到第奇数个的石头就把它往前扔规定的距离,遇到第偶数个的石头就不动它 问最远的石头据出发点(起点)的距离

优先队列的应用,想到优先队列就很好解决了

 1 #include<cstdio>
 2 #include<queue>
 3 using namespace std;
 4 struct point {
 5     int x,y;
 6     bool operator <(const point & q) const
 7     {
 8         if (x==q.x) return y>q.y;
 9         else return x>q.x;
10     }
11 };
12 int main()
13 {
14     int t,n;
15     while (~scanf("%d",&t))
16     {
17         while (t--)
18         {
19             int k=1;
20             priority_queue<point> que;
21             point temp;
22             scanf("%d",&n);
23             while (n--)
24             {
25                scanf("%d %d",&temp.x,&temp.y);
26                que.push(temp);
27             }
28             while (!que.empty())
29             {
30                 temp=que.top(),que.pop();
31                 if (k%2==1)
32                 {
33                     temp.x+=temp.y;
34                     que.push(temp);
35                 }
36                 k++;
37             }
38             printf("%d\n",temp.x);
39         }
40     }
41     return 0;
42 }