zoj2319Beautiful People Dp

最长上升子序列的nlogn的解法，不过要注意的是 x按从小到大排序，y要按从大到小排序。对于同样的x，在维护的单调数组中y应该保留较小的那个。

#include<iostream>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<vector>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int maxn = 111111;

typedef long long LL;

struct Node
{
    int x; int y; int id;
}node1[maxn],node[maxn];


int erfen(int a[], int l, int r, int key)
{
    int ans = -1;
    while (l <= r){
        int mid = (l + r) >> 1;
        if (a[mid] > key) {
            ans = mid;
        }
        if (a[mid] > key) r = mid - 1;
        else l = mid + 1;
    }
    return ans;
}

int cmp(const Node &a,const Node &b)
{
    if(a.x==b.x) return a.y>b.y;
    return a.x<b.x;
}
int ans[maxn];
int cnt;
int vis[maxn];
int father[maxn];
void showpath(int x)
{
    while(x!=father[x]){
        printf("%d ",x);x=father[x];
    }
    printf("%d",x);
    cout<<endl;
}

int main()
{
    int T,n;
    cin >> T;
    while (T--){
        cin >> n;
        for (int i = 1; i <= n; i++){
            scanf("%d%d", &node[i].x, &node[i].y);
            node[i].id = i;
            node1[i] = node[i];
        }
        sort(node + 1, node + 1 + n, cmp);
        cnt = 1;
        ans[cnt] = node[1].y; vis[cnt++] = node[1].id;
        father[node[1].id] = node[1].id;
        //cout<<endl;
        //for(int i = 1;i<=n;i++)
           // printf("%d %d
",node[i].x,node[i].y);
        //cout<<endl;
        for (int i = 2; i <= n; i++){
            int t = erfen(ans, 1, cnt - 1, node[i].y);
            //printf("%d %d %d
",t,ans[cnt-1],node[i].y);system("pause");
            if (t == -1 && node1[vis[cnt - 1]].x < node[i].x&&node1[vis[cnt-1]].y<node[i].y){
                father[node[i].id] = vis[cnt-1];
                ans[cnt] = node[i].y; vis[cnt++] = node[i].id;;
            }
            if (t != -1){
                if(t!=1&&node[i].y<=ans[t-1]) continue;
                ans[t] = node[i].y; vis[t] = node[i].id;
                if(t==1) father[node[i].id] = node[i].id;
                else father[node[i].id] = vis[t-1];
            }
        }
        printf("%d 
",cnt-1);
        showpath(vis[cnt-1]);
    }
    return 0;
}