ZOJ 2319 Beautiful People

LIS。先按S降序升序再按B降序排序（如果B不按降序排序的话就会覆盖掉正解），然后再对B用O(nlog(n))的LIS求解就可以了。用d数组标记每个元素在上升序列中的位置，然后根据d倒着找id就可以了。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<fstream>
#include<sstream>
#include<cstdlib>
#include<vector>
#include<string>
#include<cstdio>
#include<bitset>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define MP make_pair
using namespace std;

const int N = 100100;
const int INF = 1e9 + 7;

struct Node
{
    int s, b, id;
    bool operator < (const Node &rhs) const
    {
        return s < rhs.s || (s == rhs.s && b > rhs.b);
    }
}p[N];
int g[N];
int d[N];

int main()
{
    int s, b, t, n, ans;
    scanf("%d", &t);
    while(t --)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; i ++)
        {
            scanf("%d%d", &s, &b);
            p[i].s = s, p[i].b = b, p[i].id = i + 1;
        }
        sort(p, p + n);ans = 0;
        for(int i = 1; i <= n; i ++) g[i] = INF;
        for(int i = 0; i < n; i ++)
        {
            int k = lower_bound(g+1, g+n+1, p[i].b) - g;
            g[k] = p[i].b;
            d[i] = k;
            ans = max(ans, k);
        }
        printf("%d\n", ans);
        int pt = -1;
        for(int i = n - 1; i >= 0; i --)
        {
            if(p[i].s != pt && d[i] == ans)
            {
                printf("%d", p[i].id);
                if(ans) putchar(' ');
                ans --;pt = p[i].s;
            }
            if(!ans) break;
        }puts("");
        if(t) puts("");
    }
}