计蒜客 2017icpc南宁赛区 The Heaviest Non-decreasing Subsequence Problem 最长不递减子序列
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https://nanti.jisuanke.com/t/17319
题意:对于每个数x,若x<0,则价值为0,x大于10000,价值为5且数值变为x-10000,反之价值为1。现在求不递减的序列使得价值最大。
题解:负数没有贡献不考虑,大于10000的存五次,相当于价值为1,反之存一次。然后求一次最长不递减子序列即可,答案就是子序列长度,因为价值都是1,这样处理之后,没有差别了。
本来应该是道水题,网络赛的时候,最后没时间自己脑残想了个二维的dp。还有补题的时候,数据保证长度不超过2*10e5,但实际上还要大。不知道是不是放题库的时候加入了数据?
代码:
#include<set>#include<map>#include<stack>#include<queue>#include<vector>#include<string>#include<bitset>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>#include<iomanip>#include<iostream>#define debug cout<<"aaa"<<endl#define d(a) cout<<a<<endl#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define lson l,mid,root<<1#define rson mid+1,r,root<<1|1#define MIN_INT (-2147483647-1)#define MAX_INT 2147483647#define MAX_LL 9223372036854775807i64#define MIN_LL (-9223372036854775807i64-1)using namespace std;const int N = 1000000;const int mod = 1000000000 + 7;const double eps = 1e-8;int dp[N],a[N];int n,len,pos,x;//dp[i]:保存的是最长递增子序列长度为i的结尾的最小值 //可以当作往里面一个一个加数字,更新最小那个 int search(int l,int r,int x){while(l<=r){int mid=(l+r)>>1;if(dp[mid]>x){r=mid-1;}else{l=mid+1;}}return l;}void LIS(){mem(dp,0),len=1;dp[1]=a[1];for(int i=2;i<=n;i++){if(a[i]>=dp[len]){dp[++len]=a[i];}else{pos=search(1,len,a[i]);dp[pos]=a[i]; }}}int main(){n=0;while(~scanf("%d",&x)){if(x<0) continue;if(x>=10000){for(int i=1;i<=5;i++){a[++n]=x-10000;}}else{a[++n]=x;}}LIS();printf("%d\n",len);return 0;}
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