Constructing Roads In JGShining’s KingdomTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19731Accepted Submission(s): 5581
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.
题意：河岸两旁有n个村庄，他们之间要互相修路，，并且同一边的不互相修，在保证不交叉的情况下，最大限度的路的是多少。
eg： 1 23456789poor
123456789rich
其中1-2,2-3,3-1,4-6,5-5,6-4,7-9,8-7,9-8;短线表示相连，要求不重合（交叉）那么最多修5条；
思路：就是排完序后对应的另一边的位置
eg：123456789poor
231654978rich
对rich边求最长公共递增子序列！
时间复杂度O(n*log(n))
链接：?pid=1025
#include<cstdio>#include<algorithm>using namespace std;struct node{int p;int r;} num[500005];int arr[500005];int cmp(node a,node b){if(a.p==b.p) return a.r<b.r;elsereturn a.p<b.p;}int main(){int n,ca=1;while(scanf("%d",&n)!=EOF){for(int i=0; i<n; i++)scanf("%d%d",&num[i].p,&num[i].r);sort(num,num+n,cmp);/**O(n*log(n))*/arr[0]=num[0].r;int cnt=1;for(int i=1; i<n; i++){if(arr[cnt-1]<=num[i].r){if(arr[cnt-1]<num[i].r)arr[cnt++]=num[i].r;continue;}else{int left=0;int right=cnt;while(left<=right){int mid=(left+right)/2;if(arr[mid]<num[i].r) left=mid+1;else if(arr[mid]>num[i].r) right=mid-1;else{left=mid;break;}}arr[left]=num[i].r;}}printf("Case %d:\nMy king, at most %d %s can be built.\n\n",ca++,cnt,cnt!=1?"roads":"road");}return 0;}还有一个可以优化的地方，这个还是很巧的，那就是不用排序了，直接把poor村的作为下标，这样就不用排序了；又快了一倍。
#include <iostream>#define maxn 500002using namespace std;int dp[maxn];int n,t;int p[maxn];void Solve(){int i,low,up,mid,len=1;dp[1] = p[1];for(i=1;i<=n;i++){low = 1;up = len;while(low<=up){mid = (low+up)/2;if(dp[mid]>=p[i])up = mid-1;else low = mid+1;}dp[low] = p[i];if(low>len) len++;}printf("Case %d:/nMy king, at most %d road",t++,len);if(len!=1) printf("s");printf(" can be built./n/n");}int main(){int i,temp,r;t = 1;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++){scanf("%d%d",&temp,&r);p[temp]=r;}Solve();}return 0;}
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