uva 11136 - Hoax or what

用两个优先队列来实现,因为队列只能从一头出去;

所以维护一个数组,来标记这个队列的已经出列而另外一个队列没有出列的元素;

到时候再把他们删了就行;

 1 #include<cstdio>
 2 #include<queue>
 3 #include<cstring>
 4 #define maxn 1000009
 5 using namespace std;
 6 
 7 priority_queue<int,vector<int>,greater<int> >gq;
 8 priority_queue<int,vector<int>,less<int> >lq;
 9 int numg[maxn],numl[maxn];
10 int main()
11 {
12     int n,x,k;
13     while(scanf("%d",&n)&&n)
14     {
15         memset(numl,0,sizeof numl);
16         memset(numg,0,sizeof numg);
17         long long ans=0;
18         while(!gq.empty())gq.pop();
19         while(!lq.empty())lq.pop();
20         for(int i=0;i<n;i++)
21         {
22             scanf("%d",&k);
23             while(k--)
24             {
25                 scanf("%d",&x);
26                 gq.push(x);
27                 lq.push(x);
28             }
29             while(1)
30             {
31                 if(numg[lq.top()]>0)
32                 {
33                     numg[lq.top()]--;
34                     lq.pop();
35                 }
36                 else break;
37             }
38             while(1)
39             {
40                 if(numl[gq.top()]>0)
41                 {
42                     numl[gq.top()]--;
43                     gq.pop();
44                 }
45                 else break;
46             }
47             numg[gq.top()]++;
48             numl[lq.top()]++;
49             ans+=lq.top()-gq.top();
50             gq.pop();
51             lq.pop();
52         }
53         printf("%lld\n",ans);
54     }
55     return 0;
56 }
View Code

还可以用multiset来做;

代码:

 1 #include<cstdio>
 2 #include<set>
 3 using namespace std;
 4 
 5 multiset<int>s;
 6 
 7 int main()
 8 {
 9     int n,k,x;
10     while(scanf("%d",&n)&&n)
11     {
12         long long ans=0;
13         s.clear();
14         while(n--)
15         {
16             scanf("%d",&k);
17             while(k--)
18             {
19                 scanf("%d",&x);
20                 s.insert(x);
21             }
22             multiset<int>::iterator it;
23             it=s.begin();
24             int mi=*it;
25             s.erase(it);
26             it=s.end();
27             it--;
28             int ma=*it;
29             s.erase(it);
30             ans+=ma-mi;
31         }
32         printf("%lld\n",ans);
33     }
34     return 0;
35 }
View Code

 

 

posted @ 2013-10-24 17:28  Yours1103  阅读(200)  评论(0编辑  收藏  举报