hdu1025 LIS

题目链接：点击打开链接

题目是给你匹配对，第i个穷人城市需要第j个富人城市的帮助，穷人与富人城市是平行线，要求不能有交叉，比如1—>2,2->1是不行的，求最多能被帮助的城市数量。

思路：可看成LIS问题。

代码是别人的模板代码

#include <iostream>#include<cstdio>using namespace std;const int maxn=500000+10;int Arr[maxn],List[maxn];int Stack[maxn];int LIS(int *Arr,int N){    int top = 0;    Stack[top] = -1;    for(int i = 1; i <= N; ++i)    {        if(Arr[i] > Stack[top])            Stack[++top] = Arr[i];        else        {            int low = 1;            int high = top;            while(low <= high)            {                int mid = (low + high)/2;                if(Arr[i] > Stack[mid])                    low = mid + 1;                else                    high = mid - 1;            }            Stack[low] = Arr[i];        }    }    return top;}int main(){  int ans,i,j,n,a,b;  ans=0;  while(~scanf("%d",&n))  {      for(i=0;i<n;i++)      {            scanf("%d %d",&a,&b);           Arr[a]=b;      }      int k=LIS(Arr,n);      if(k==1)    printf("Case %d:\nMy king, at most %d road can be built.\n\n",++ans,k);    else  printf("Case %d:\nMy king, at most %d roads can be built.\n\n",++ans,k);  }    return 0;}