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B：对于坐标先进行离散化，然后线性的扫一遍，常见的处理方式。

lines

2
5
1 2 
2 2
2 4
3 4
5 1000
5
1 1
2 2
3 3
4 4
5 5

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#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <math.h>
#include <time.h>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
///#define LL long long
#define LL __int64
#define INF 0x3f3f3f
#define PI acos(-1)
#define mod 1000000007


using namespace std;


const int maxn = 200010;

int num[maxn];
int dx[maxn];
int lx[maxn];
int ly[maxn];

struct node
{
    int x, y;
}f[maxn];

int Find(int x, int n)
{
    int l = 0;
    int r = n-1;
    while(l <= r)
    {
        int mid = (l+r)/2;
        if(dx[mid] == x) return mid;
        if(dx[mid] > x) r = mid-1;
        else l = mid+1;
    }
    return -1;

}

int main()
{
    int T;
    cin >>T;
    while(T--)
    {
        int n;
        int ans = 0;
        scanf("%d",&n);
        for(int i = 0; i < n; i++)
        {
            scanf("%d %d",&f[i].x, &f[i].y);
            num[ans++] = f[i].x;
            num[ans++] = f[i].y;
        }
        sort(num, num+ans);
        int cnt = 0;
        dx[cnt++] = num[0];
        for(int i = 1; i < ans; i++)
        {
            if(num[i] == dx[cnt-1]) continue;
            dx[cnt++] = num[i];
        }
        memset(lx, 0, sizeof(lx));
        memset(ly, 0, sizeof(ly));
        for(int i = 0; i < n; i++)
        {
            int x = Find(f[i].x, cnt);
            int y = Find(f[i].y, cnt);
            lx[x]++;
            ly[y]--;
        }
        int Max = 0;
        int sum = 0;
        for(int i = 0; i < cnt; i++)
        {
            sum += lx[i];
            Max = max(sum, Max);
            sum += ly[i];
        }
        cout<<Max<<endl;
    }
    return 0;
}

直接想的话dp(i, j, k)已经交换了i次，位置j处，取A(k == 0)或者取B(k == 1)时的最大长度为多少。

三层会1000*1000*1000TLE到死啊。

有m颗线段树维护每次找到交换了i次位置为j的最长子序列是多少。

magic balls

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5 1
4 2
3 1
2 4
3 1
5 4
5 1
4 2
3 1
2 4
3 1

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#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define mod 1000000007
#define rson mid+1, r, rt<<1|1
#define lson l, mid, rt<<1
#define root 0, ans-1, 1


const int maxn = 1001;

using  namespace std;

int dp[maxn][8001];
int dy[2*maxn], dx[2*maxn];
int num1[maxn], num2[maxn];

struct node
{
    int x, y;
}f[maxn];
int ans;

int Find(int x)
{
    int l = 0;
    int r = ans-1;
    while(l <= r)
    {
        int mid = (l+r)>>1;
        if(dy[mid] == x) return mid;
        else if(dy[mid] < x) l = mid+1;
        else r = mid-1;
    }
    return -1;
}


int Query(int m, int l, int r, int L, int R, int site)
{
    if(l == L && r == R) return dp[m][site];
    int mid = (L+R)>>1;
    if(r <= mid) return Query(m, l, r, L, mid, site<<1);
    else if(l > mid) return Query(m, l, r, mid+1, R, site<<1|1);
    else return max(Query(m, l, mid, L, mid, site<<1), Query(m, mid+1, r, mid+1, R, site<<1|1));
}

void Push_UP(int m, int site)
{
    dp[m][site] = max(dp[m][site<<1], dp[m][site<<1|1]);
}

void Updata(int m, int x, int d, int l, int r, int site)
{
    if(l == r)
    {
        dp[m][site] = d;
        return;
    }
    int mid = (l+r)>>1;
    if(x <= mid) Updata(m, x, d, l, mid, site<<1);
    else Updata(m, x, d, mid+1, r, site<<1|1);
    Push_UP(m, site);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n, m;
        scanf("%d %d",&n, &m);
        ans = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d %d",&f[i].x, &f[i].y);
            dx[i<<1] = f[i].x;
            dx[i<<1|1] = f[i].y;
        }
        sort(dx, dx+2*n);
        dy[ans++] = dx[0];
        for(int i = 1; i < 2*n; i++)
        {
            if(dy[ans-1] == dx[i]) continue;
            dy[ans++] = dx[i];
        }
        for(int i = 0; i <= m ; i++)
            for(int j = 0; j <= 4*ans; j++) dp[i][j] = 0;
        int Max = 0;
        int sum1, sum2, xans;
        for(int i = 0; i < n; i++)
        {
            int x = Find(f[i].x);
            int y = Find(f[i].y);
            xans = 0;
            if(x) xans = Query(0, 0, x-1, 0, ans-1, 1);
            num1[0] = xans+1;
            Max = max(Max, xans+1);
            for(int j = 1; j <= i+1 && j <= m; j++)
            {
                sum1 = 0;
                sum2 = 0;
                if(j != i+1)
                {
                    if(x) sum1 = Query(j, 0, x-1, 0, ans-1, 1);
                    num1[j] = sum1+1;
                }
                if(y) sum2 = Query(j-1, 0, y-1, 0, ans-1, 1);
                num2[j] = sum2+1;
                xans = max(sum1, sum2);
                Max = max(Max, xans+1);
            }
            for(int j = 0; j <= i+1 && j <= m; j++)
            {
                if(j != i+1) Updata(j, x, num1[j], 0, ans-1, 1);
                if(!j) continue;
                Updata(j, y, num2[j], 0, ans-1, 1);
            }
        }
        printf("%d\n",Max);
    }
    return 0;
}

BestCoder Round #20 B,C

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原文地址：http://blog.csdn.net/xu12110501127/article/details/41673775