LIS&LICS

来源：互联网发布：淘宝人肉一个人多钱编辑：程序博客网时间：2024/06/03 20:16

定义d[k]:长度为k的上升子序列的最末元素，若有多个长度为k的上升子序列，则记录最小的那个最末元素。
注意d中元素是单调递增的，下面要用到这个性质。
首先len = 1,d[1] = a[1],然后对a[i]:若a[i]>d[len]，那么len++，d[len] = a[i];
否则，我们要从d[1]到d[len-1]中找到一个j，满足d[j-1]

#include <stdio.h>#include <string.h>#include <iostream>int a[1010];int d[1010];int kill(int len, int x){    int b = 1, e = len;    while(b <= e)    {        int mid = (b + e) / 2;        if(x > d[mid])        {            b = mid + 1;        }        else{            e = mid - 1;        }    }    return b;}int main(){    int n;    while(!scanf("%d", &n))    {        for(int i = 1; i <= n; i++)        {            scanf("%d", &a[i]);        }        d[1] = a[1];        int len = 1;        int j;        for(int i = 2; i <= n; i++)        {            if(d[1] >= a[i])            {                j = 1;            }            else if(a[i] > d[len])            {                j = ++len;            }            else{                j = kill(len, a[i]);            }            d[j] = a[i];        }        printf("%d\n", len);    }    return 0;}

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;const int MAXN = 1010;int n;int a[MAXN];int dp[MAXN];int lis(){    memset(dp, 0, sizeof(int)*n);    int len = 1;    dp[0] = a[0];    for (int i = 1; i < n; ++i)    {        int pos = lower_bound(dp, dp + len, a[i]) - dp;        dp[pos] = a[i];      //  printf("%d %d\n",dp[pos], len);        len = max(len, pos + 1);    }    return len;} int main() {     scanf("%d", &n);     for(int i = 0; i < n; i++)     {         scanf("%d", &a[i]);     }     //printf("%d\n",  lis());     int len = lis();     for(int i = 0; i < len; i++)     {         printf("%d ", dp[i]);     } }

LICS

#include <string.h>#include <stdio.h>#include <iostream>using namespace std;int f[1005][1005], a[1005], b[1005], i, j, t, n1, n2, maxn;int main(){    scanf("%d%d", &n1, &n2);    for(i =0; i <= n1; i++)    {        scanf("%d", &a[i]);    }    for(i = 1; i <= n2; i++)    {        scanf("%d", &b[i]);    }    memset(f, 0, sizeof(f));    for(i = 1; i <= n1; i++)    {        maxn = 0;        for(j = 1; j <= n2; j++)        {            f[i][j] = f[i - 1][j];            if(a[i] > b[j] && maxn < f[i - 1][j])            {                maxn = f[i - 1][j];            }            if(a[i] == b[j])            {                f[i][j] = maxn + 1;            }        }    }    maxn = 0;    for(i = 1; i <= n2; i++)    {        if(maxn < f[n1][i])        {            maxn = f[n1][i];        }    }    printf("%d\n", maxn);}

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