See LCS again

See LCS again

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

There are A, B two sequences, the number of elements in the sequence is n、m;

Each element in the sequence are different and less than 100000.

Calculate the length of the longest common subsequence of A and B.

输入

The input has multicases.Each test case consists of three lines;
The first line consist two integers n, m (1 < = n, m < = 100000);
The second line with n integers, expressed sequence A;
The third line with m integers, expressed sequence B;

输出

For each set of test cases, output the length of the longest common subsequence of A and B, in a single line.

样例输入

5 41 2 6 5 41 3 5 4

样例输出

3

//无语了，什么都不想说。。。#include <stdio.h>#include <string.h>int mark[100001], data[100001];int binarySearch(int key, int n)     //二分查找大于等于且距离key最近的数的位置{    int left = 0, right = n, mid;    while(left <= right)    {        mid = (left + right) / 2;        if(mark[mid] >= key)        {            right = mid - 1;        }        else        {            left = mid + 1;        }    }    return left;}/*void fun()       //测试二分查找是否正确{    int i, n, temp;    scanf("%d", &n);    for(i = 0; i < n; i++)    {        scanf("%d", &data[i]);    }    while(scanf("%d", &temp) != EOF)    {        printf("%d\n", binarySearch(temp, n-1));    }}*/int main(){   // fun();    int n, m, i, temp, top, ans;    while(~scanf("%d%d", &n, &m))    {        memset(mark, 0, sizeof(mark));        top = 0;        for(i = 1; i <= n; i++)        {            scanf("%d", &temp);            mark[temp] = i;              //将原序列单调上升编号，并标记每个数        }        for(i = 1; i <= m; i++)        {            scanf("%d", &temp);            if(mark[temp])            {                data[top++] = mark[temp];   //将两个序列中都有的数，以在第二个序列中的相对位置存入新序列            }        }        ans = 0;        mark[ans++] = data[0];        for(i = 1; i < top; i++)       //判断单调最长子序列        {            if(data[i] > mark[ans-1])            {                mark[ans++] = data[i];            }            else            {                mark[binarySearch(data[i], ans-1)] = data[i];            }        }        printf("%d\n", ans);    }    return 0;}