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_pairusing 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("");    }}