HDU:1025 Constructing Roads In JGShining's Kingdom(LIS-n*logn解法+思维)

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Constructing Roads In JGShining's Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23170    Accepted Submission(s): 6619


Problem Description
JGShining's kingdom consists of 2n(n is no more than 500,000) small cities which are located in two parallel lines.

Half of these cities are rich in resource (we call them rich cities) while the others are short of resource (we call them poor cities). Each poor city is short of exactly one kind of resource and also each rich city is rich in exactly one kind of resource. You may assume no two poor cities are short of one same kind of resource and no two rich cities are rich in one same kind of resource. 

With the development of industry, poor cities wanna import resource from rich ones. The roads existed are so small that they're unable to ensure the heavy trucks, so new roads should be built. The poor cities strongly BS each other, so are the rich ones. Poor cities don't wanna build a road with other poor ones, and rich ones also can't abide sharing an end of road with other rich ones. Because of economic benefit, any rich city will be willing to export resource to any poor one.

Rich citis marked from 1 to n are located in Line I and poor ones marked from 1 to n are located in Line II. 

The location of Rich City 1 is on the left of all other cities, Rich City 2 is on the left of all other cities excluding Rich City 1, Rich City 3 is on the right of Rich City 1 and Rich City 2 but on the left of all other cities ... And so as the poor ones. 

But as you know, two crossed roads may cause a lot of traffic accident so JGShining has established a law to forbid constructing crossed roads.

For example, the roads in Figure I are forbidden.



In order to build as many roads as possible, the young and handsome king of the kingdom - JGShining needs your help, please help him. ^_^
 

Input
Each test case will begin with a line containing an integer n(1 ≤ n ≤ 500,000). Then n lines follow. Each line contains two integers p and r which represents that Poor City p needs to import resources from Rich City r. Process to the end of file.
 

Output
For each test case, output the result in the form of sample. 
You should tell JGShining what's the maximal number of road(s) can be built. 
 

Sample Input
21 22 131 22 33 1
 

Sample Output
Case 1:My king, at most 1 road can be built.Case 2:My king, at most 2 roads can be built.
Hint
Huge input, scanf is recommended.
 

Author
JGShining(极光炫影)
 

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题目大意:穷国家要跟富国家之间修建公路来使得贸易流通,一个穷国家只能连接一个富国家,并且穷国家之间、富国家之间互不修路,问修多少条路能使得较多的穷国家能够流通贸易(且根据题目中所述,修建的路不能交叉)。




解题思路:









如上图所示,从穷国家一号开始向富国家建路,往后依次建路,发现后面的穷国家只能连接已经连接的富国家之后的富国家,即他连接的富国家要比前面的富国家编号大,由此,看出是求最长子序列长度的题,但是题中数据较大50W,用LIS的贪心二分法求。




代码如下:

#include <cstdio>int a[500010];int d[500010];int ans;int erfen(int x){int l=1,r=ans;int pos,mid;while(l<=r){mid=(l+r)/2;if(d[mid]>a[x]){pos=mid;r=mid-1;}else{l=mid+1;}}return pos;}int main(){int n;int k=1;while(scanf("%d",&n)!=EOF){for(int i=0;i<n;i++){int x,y;scanf("%d%d",&x,&y);a[x]=y;//这样储存的方便之处,好好体会下吧}d[1]=a[1];ans=1;for(int i=2;i<=n;i++)//从编号低的穷国家开始扫{if(d[ans]<a[i]){ans++;d[ans]=a[i];}else{d[erfen(i)]=a[i];}}if(ans==1){printf("Case %d:\n",k++);printf("My king, at most 1 road can be built.\n\n");//注意是双换行 }else{printf("Case %d:\n",k++);printf("My king, at most %d roads can be built.\n\n",ans);//注意是双换行 }}return 0;}


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