Constructing Roads In JGShining's Kingdom（最長單調遞增子序列應用+hdu1025）

 

Constructing Roads In JGShining's Kingdom

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19731 Accepted 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

2
1 2
2 1
3
1 2
2 3
3 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.
 

 

 

題意：河岸兩旁有n個村莊，他們之間要互相修路，並且同一邊的不互相修，在保證不交叉的情況下，最大限度的路的是多少。

eg： 1 2 3 4 5 6 7 8 9 poor

1 2 3 4 5 6 7 8 9 rich

其中1-2,2-3,3-1,4-6,5-5,6-4,7-9,8-7,9-8;短線表示相連，要求不重合（交叉）那麼最多修5條；

思路：就是排完序後對應的另一邊的位置

eg： 1 2 3 4 5 6 7 8 9 poor

2 3 1 6 5 4 9 7 8 rich

對rich邊求最長公共遞增子序列！

 

時間復雜度O(n*log(n))

 

 

#include
#include
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.rnum[i].r) right=mid-1;
                    else
                    {
                        left=mid;
                        break;
                    }
                }
                arr[left]=num[i].r;
            }

        }
        printf(Case %d:
My king, at most %d %s can be built.

,ca++,cnt,cnt!=1?roads:road);
    }
    return 0;
}

還有一個可以優化的地方，這個還是很巧的，那就是不用排序了，直接把poor村的作為下標，這樣就不用排序了；又快了一倍。

 

 

#include 
#define maxn 500002
using 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;
}