题意：对于给定的物品，求两个在高度上单调不递增，权值上单调不递减的序列，使二者长度之和最大。
分析：可以用费用流求解，因为要求长度和最大，视作从源点出发的流量为2的费用流，建负权边，每个物品只能取一次，且花费为-1。将每个物品拆成入点和出点，中间建容量为1，费用为-1的弧。建源点s和超级源点S，S到s建容量为2，费用为0的弧，表示只有两个序列。源点s向每个入点建容量为1，费用为0的弧，表示每个点都可作为序列的首项。出点向汇点建容量为1，费用为0的弧，表示每个点都可作为序列的末项。
对给定物品按高度和权值排序后，从权值较小的物品向权值较大的物品建边，容量为1，花费为0。
跑出费用流后对花费取反就是答案。spfa要用栈优化，队列会T。
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
const int MAXN = 2005;
const int MAXM = 2000005;
const int INF = 0x3f3f3f3f;
struct Edge{
int to, next, cap, flow, cost;
} edge[MAXM];
int head[MAXN], tot;
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N;
void init(int n)
{
N = n;
tot = 0;
memset(head, -1, sizeof(head));
}
void AddEdge(int u, int v, int cap, int cost)
{
edge[tot] = (Edge){v,head[u],cap,0,cost};
head[u] = tot++;
edge[tot] = (Edge){u,head[v],0,0,-cost};
head[v] = tot++;
}
bool spfa(int s, int t){
stack<int> q;
for (int i = 0; i < N; i++){
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while (!q.empty()){
int u = q.top();
q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next){
int v = edge[i].to;
if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost){
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if (!vis[v]){
vis[v] = true;
q.push(v);
}
}
}
}
if (pre[t] == -1) return false;
else return true;
}
int minCostMaxflow(int s, int t, int &cost){
int flow = 0;
cost = 0;
while (spfa(s, t)){
int Min = INF;
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]){
if (Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]){
edge[i].flow += Min;
edge[i ^ 1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}
struct Node{
int h,w;
bool operator<(const Node &rhs) const{
if(h==rhs.h) return w<rhs.w;
return h>rhs.h;
}
}vz[MAXN];
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int T; scanf("%d",&T);
while(T--){
int N; scanf("%d",&N);
int u,v;
for(int i=1;i<=N;++i){
scanf("%d %d",&vz[i].h,&vz[i].w);
}
sort(vz+1,vz+N+1);
init(2*N+4);
int s = 2*N+1, t = 2*N+2;
int S = 0;
for(int i=1;i<=N;++i){
AddEdge(i+N,t,1,0);
AddEdge(s,i,1,0);
AddEdge(i,i+N,1,-1);
int x = INF;
for(int j=i+1;j<=N;++j){
if(vz[i].w<=vz[j].w){
AddEdge(i+N,j,1,0);
}
}
}
AddEdge(S,s,2,0);
int cost;
minCostMaxflow(S,t,cost);
printf("%d\n",-cost);
}
return 0;
}