HDU5087 - Revenge of LIS II (次长上升子序列)

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题目链接 HDU5087

【题意】求第二长的LIS的长度。

【分析】有两种方法；

1.直接添加一个dp[]在更新最大值的同时更新第二大值就可以了。

2.用cnt[]记录每个位置到达的次数，最终统计一遍最长长度出现的次数和，只出现一次的话次长序列就是最大序列-1，否则次长序列==最长序列

【AC CODE(1)】

#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <queue>#include <stack>#include <vector>#include <string>#include <algorithm>using namespace std;#define rep(i,a,n) for(int i = a; i < n; i++)#define repe(i,a,n) for(int i = a; i <= n; i++)#define per(i,n,a) for(int i = n; i >= a; i--)#define clc(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3ftypedef long long LL;#define MAXN 1010int a[MAXN], dp[MAXN][2], n;//dp[][0]表示最大值,dp[][1]是次大值int sloved(){    dp[0][0] = 1,dp[0][1] = 0;    int fir = 1, sec = 0;    for (int i = 1; i < n; i++)    {        dp[i][0] = 1, dp[i][1] = 0;        for (int j = 0; j < i; j++)        {            if(a[j] >= a[i]) continue;            if(dp[i][0] <= dp[j][0]+1)//当前>=最大值则更新最大和次大值            {                dp[i][1] = max(dp[i][0], dp[j][1]+1);//次大值可能来自于原来次大值+1或者原来的最大值                dp[i][0] = dp[j][0]+1;            }            else if(dp[i][1] < dp[j][0]+1)//没有大于最大值，但可能大于次大值                dp[i][1] = dp[j][0]+1;        }        if(fir <= dp[i][0])        {            sec = max(dp[i][1],fir);            fir = dp[i][0];        }        else if(sec < dp[i][0])            sec = dp[i][0];    }    return sec;}int main(){#ifdef SHY    freopen("e:\\1.txt", "r", stdin);#endif    int t;    scanf("%d%*c", &t);    while(t--)    {        scanf("%d%*c", &n);        rep(i,0,n) scanf("%d%*c", &a[i]);        int s;        printf("%d\n", sloved());    }    return 0;}

【AC CODE(2)】

#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <queue>#include <stack>#include <vector>#include <string>#include <algorithm>using namespace std;#define rep(i,a,n) for(int i = a; i < n; i++)#define repe(i,a,n) for(int i = a; i <= n; i++)#define per(i,n,a) for(int i = n; i >= a; i--)#define clc(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3ftypedef long long LL;#define MAXN 1010int a[MAXN], dp[MAXN], n, cnt[MAXN];int sloved(){    dp[0] = cnt[0] = 1;    int ans = 1, s = 0;    for (int i = 1; i < n; i++)    {        dp[i] = cnt[i] = 1;        for (int j = 0; j < i; j++)        {            if(a[j] >= a[i]) continue;            if(dp[i] < dp[j]+1)                dp[i] = dp[j]+1, cnt[i] = cnt[j];            else if(dp[i] == dp[j]+1)                cnt[i] += cnt[j];        }        if(ans < dp[i])            ans = dp[i], s = i;    }    int sum = 0;    rep(i,0,n) if(ans == dp[i]) sum += cnt[i];    if(1 == sum) ans--;    return ans;}int main(){#ifdef SHY    freopen("e:\\1.txt", "r", stdin);#endif    int t;    scanf("%d%*c", &t);    while(t--)    {        scanf("%d%*c", &n);        rep(i,0,n) scanf("%d%*c", &a[i]);        printf("%d\n", sloved());    }    return 0;}

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