3831: [Poi2014]Little Bird Time Limit: 20 Sec  Memory Limit: 128 MBSubmit: 505  Solved: 322[Submit][Status][Discuss] Description In the Byteotian Line Forest there are   trees in a row. On top of the first one, there is a little bird who would like t…

设f[i]表示到i最少休息次数,f[i]=min(f[j]+(h[j]<=a[i])),i-k<=j<i,单调队列优化DP #include<cstdio> #define N 1000010 inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<…

f(i)=min{f(j)+(D(j)<=D(i))} (max(1,i-k)<=j<=i) 有两个变量,很难用单调队列,但是(引用): 如果fi<fj,i一定比j优秀.因为如果heighti≥heightj就不用说了,如果heighti<heightj,i可以花1的代价跳到高的地方,顶多和j一样,不会比j差. 如果fi=fj,那当然就是谁的height高谁优秀. 双关键字搞一下就好了. #include<cstdio> using namespace std;…

[BZOJ3831][Poi2014]Little Bird Description In the Byteotian Line Forest there are   trees in a row. On top of the first one, there is a little bird who would like to fly over to the top of the last tree. Being in fact very little, the bird might lack…

3831: [Poi2014]Little Bird Time Limit: 20 Sec Memory Limit: 128 MB Submit: 310 Solved: 186 [Submit][Status][Discuss] Description In the Byteotian Line Forest there are trees in a row. On top of the first one, there is a little bird who would like to…

P3572 [POI2014]PTA-Little Bird 一只鸟从1跳到n.从1开始,跳到比当前矮的不消耗体力,否则消耗一点体力,每次询问有一个步伐限制k,求每次最少耗费多少体力 很简短的题目哼. 首先对于一个点, 他的状态一定是由前 \(k\) 个转移过来的. \(k\) 的长度在每组询问内一定, 想到用单调队列维护 \(dp\) . 不过此时单调队列里的元素有两个关键字: 劳累度和高度, 因为跳到比这个点高的树需要花费恒为一点体力(这个很重要), 所以我们维护单调队列的时候可以以劳累度为…

P3572 [POI2014]PTA-Little Bird 题目描述 In the Byteotian Line Forest there are nn trees in a row. On top of the first one, there is a little bird who would like to fly over to the top of the last tree. Being in fact very little, the bird might lack the s…

P3572 [POI2014]PTA-Little Bird 题目描述 In the Byteotian Line Forest there are nn trees in a row. On top of the first one, there is a little bird who would like to fly over to the top of the last tree. Being in fact very little, the bird might lack the s…

题面:[POI2014]PTA-Little Bird 题解: N<=1e6 Q<=25F[i]表示到达第i棵树时需要消耗的最小体力值F[i]=min(F[i],F[j]+(D[j]>=D[i])) (j>=i-K)使用单调队列维护越小的越优,在写单调队列时,让F值最小的数越前因为F[i]-F[j]最多等于1然后如果F值相同,则D越大的越优,因为D越大,后面不用+1的概率越大大概就是这样 代码: #include<cstdio> #include<cstring&…

P3572 [POI2014]PTA-Little Bird 首先,这道题的暴力dp非常好写 就是枚举所有能转移到他的点,如果当前枚举到的位置的值大于 当前位置的话,\(f[i]=min(f[i],f[j])\) 否则就\(f[i]=min(f[i],f[j]+1)\) 时间复杂度为O(nk) 考虑优化,因为只是加1或不加,实际上只需要知道能转移到当前节点的位置 的&f&数组最小值即可,同时也要选这个位置数值最大的(想一想为什么) 这样的话,这道题就没了 #include<iostr…