hdu 1025 Constructing Roads In JGShining's Kingdom（最长上升序列n*logn算法）

用线段树实现求最长上升子序列，把复杂度降低到了n*log(n)*2。

#include<stdio.h>#include<string.h>#define N 500005struct node{    int x,y;    int max;} a[N*3];int b[N];int Max(int x,int y){    if(x>y)        return x;    return y;}void CreatTree(int t,int x,int y){    a[t].x=x;    a[t].y=y;    a[t].max=0;    if(x==y)        return ;    int mid=(x+y)/2;    int temp=t*2;    CreatTree(temp,x,mid);    CreatTree(temp+1,mid+1,y);    return ;}void InsertTree(int t,int x,int k){    if(a[t].x==a[t].y&&a[t].x==x)    {        a[t].max=k;        return ;    }    int mid=(a[t].x+a[t].y)/2;    int temp=t*2;    if(x<=mid)        InsertTree(temp,x,k);    else        InsertTree(temp+1,x,k);    a[t].max=Max(a[temp].max,a[temp+1].max);    return ;}int FindTree(int t,int x,int y){    if(a[t].x==x&&a[t].y==y)    {        return a[t].max;    }    int temp=t*2;    int mid=(a[t].x+a[t].y)/2;    if(y<=mid)        return FindTree(temp,x,y);    else if(x>mid)        return FindTree(temp+1,x,y);    else        return Max(FindTree(temp,x,mid),FindTree(temp+1,mid+1,y));}int main(){    int n;    int cnt=1;    while(scanf("%d",&n)!=EOF)    {        int i,x,y;        for(i=1; i<=n; i++)        {            scanf("%d%d",&x,&y);            b[x]=y;        }        CreatTree(1,1,n);        for(i=1; i<=n; i++)        {            int temp;            temp=FindTree(1,1,b[i])+1;            InsertTree(1,b[i],temp);        }        if(a[1].max==1)        {            printf("Case %d:\n",cnt++);            printf("My king, at most %d road can be built.\n\n",a[1].max);        }        else        {            printf("Case %d:\n",cnt++);            printf("My king, at most %d roads can be built.\n\n",a[1].max);        }    }    return 0;}

再补充一个之前看到过的别人的算法，通过维护一个上升的数组，使用二分查找，将时间复杂度降低到了n*logn。

#include<stdio.h>#include<string.h>#define N 500005int a[N],dp[N];int main(){    int n,cnt;    cnt=1;    while(scanf("%d",&n)!=EOF)    {        int x,y;        int i;        for(i=1;i<=n;i++)        {            scanf("%d%d",&x,&y);            a[x]=y;        }        int l,r,mid;        int len;        dp[1]=a[1];len=1;        for(i=2;i<=n;i++)        {            l=1;r=len;            while(l<=r)            {                mid=(l+r)/2;                if(a[i]>dp[mid]) l=mid+1;                else r=mid-1;            }            dp[l]=a[i];            if(l>len) len++;        }        if(len==1)        {            printf("Case %d:\n",cnt++);            printf("My king, at most %d road can be built.\n\n",len);        }        else        {            printf("Case %d:\n",cnt++);            printf("My king, at most %d roads can be built.\n\n",len);        }    }    return 0;}