optimal idm

Pre-Knowledge:　　path : A path from one node in the tree to another, and the number of branches on the path becomes the length of the route;the path length of the tree : the sum of the length of the path from the root to each leaf;　　The belt-weighted path length of a node: the product of the path length from the node to the root of the tree and the value of the node.　　the length of the weighted path of the tree : The sum of the weighted path length of all leaf nodes;Any two binary trees with the

is a simple construction problem.Please observe the formula:Inside the absolute value is |di-(n-i) |, which is the absolute value of the difference between Di and (N-i).In fact, for any n, we can construct a difference of 0 for each di of I and (n-i).In other words, this minimum value must be constructed to be 0.Suppose the input is 6:So you can construct this: 1 3 5 5 3 1 2 4 6 4 2 6Suppose the input is 7:So you can construct this: 1 3 5 7 5 3 1 2 4 6 6 4 2 7From the above can be seen how to co

the minimum value#include"iostream"#include"limits.h"#include"algorithm"#defineN 1001using namespacestd;intGraph[n][n];//Asumme index start from 1intV, E;//vertex number, edge numberintMinidx (intKey[],BOOLmstset[]) { intmin = Int_max, Min_idx =-1; for(intI=1; i) { if(!mstset[i] Key[i] min) {min=Key[i]; Min_idx=i; } } returnMin_idx;}intPrimmst () {intKey[n]; for(intI=1; iInt_max; BOOLMstset[n];//indicate wether in minimum spanning tree set for(intI=1; ifalse; key[1] =0; intCost =0;

Simple interval DP.#include #include#include#include#include#include#include#includeusing namespacestd;Const intMAXN = -;intN;intA[MAXN], SUM[MAXN], DP[MAXN][MAXN];intMain () { while(~SCANF ("%d", N)) {sum[0] =0; for(inti =1; I "%d", A[i]); for(inti =1; I 1] +A[i]; Memset (DP,0,sizeofDP); for(inti =2; I ) { for(intj =1; j+i-1) { intst = j, en = j + I-1; intAns =0x7FFFFFFF; for(intK = st; K ) ans= Min (Dp[st][k-1] + dp[k +1][en] + sum[en]-sum[st-1]

Transmission DoorMinshen Talk about network flow application examples, to water a waterTo write this problem, you need to understand two concepts, XOR and minimal cut.Different or have relative independence, so we take each one to see, that is, do about $32$ times the smallest cut. Then accumulate.And then the smallest cut splits a graph into two sets, simply looking at 0 sets and 1 sets.Simple diagram:The original is changed to two-way edge, all traffic limit is 1. Then all S point to point 1 o

Test instructions: Input t,t group sample, each set of sample input N, V, K. then enter the value of n items, and then enter the volume of n items. The solution of K-optimization is obtained.Analysis: Dp[n][v][k] represents n items, in the case of a volume not exceeding V, the value of the K-large is how much. DP[I][V][K] and Dp[i-1][v][k] and dp[i-1][v-volume[i]]+value[i].#include #include#include#include#includeusing namespacestd;Const intMAXN = the;intVALUE[MAXN];intVOLUME[MAXN];intdp[maxn*Te

) {ans = mid; High = mid-1; }Else{low = mid+1; } }printf("%d\ n", ans); }return 0;}Adjacency matrix procedure, slightly slower than adjacency table#include #include #include #include #define NConstintINF =0x3f3f3f3f; using namespace Std;intK, C,m, A[n][n], St, EN, N;intFirst[n], Level[n],Q[n];intCap[n][n];void Init (intdis) {memset (Cap,0, sizeof (CAP)); n = k+c; St =0; En = n+1; for(inti =1; I 0][i] =m; } for(inti = k +1; I 1; } for(inti =1; I for(intj = k +1; J if(A[i

requires a front and back two boundaries. Use M[i][j] to represent the minimum number of operations used by the product AI....J. Assuming that the last multiplication occurred between AI.....K and a (k+1) .... J, then M[i][j]=min (i≤k Below is the C language implementation of the Code #include #include#includestring.h>using namespacestd;Const intmax=10000;intm[101][max];intdpint*p,intN) { for(intR=2; r){ for(intI=1; i1; i++){ intj=i+r-1; M[I][J]=m[i+1][j

Ideas: The X-axis and the y-axis of the case is independent, can be considered separately, so long as the enumeration position, can maintain the minimum value can be, as long as the formula is taken apart, preprocessing two prefixes and can, one is the prefix of W and a w * x prefix andCode:#include Bnuoj 13268 Aladdin and the Optimal invitation

Optimal irrigation Time limit: 1.0s Memory Limit: 256.0MB Problem DescriptionRay Ray contracted a lot of wheat fields, in order to irrigate the wheat fields, Ray Ray dug a very deep well in the first wheat field, all the wheat fields from this well to water diversion irrigation.In order to irrigate, Ray Ray need to establish some canals to connect wells and wheat fields, and Ray Ray can also use some

Uva_10304 At first, I did not understand the characteristics of the Binary Search Tree. The left subtree node of the Binary Search Tree of this question must be smaller than the current node, and the right subtree node must be larger than the current node. In this way, if a root node is selected for the linear table given by the question, the left of the root node must be the left subtree, And the right must be the right subtree. This is also true for every subtree. So we take F [I] [J] to

, telling the three-dimensional coordinates of each village, the distance between the villages in the horizontal direction, the height difference in the vertical direction, and the minimum cost and length of the village.Ideas: Classic 01 Score planning questions, refer to this great God's explanation should be able to understand: http://www.cnblogs.com/Fatedayt/archive/2012/03/05/2380888.htmlCode:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permissio

=edges[i].nxt) {Edgee =Edges[i]; if(E.cap d[e.v] > d[u]+e.cost) {D[E.V]= D[u] +E.cost; P[E.V]=i; A[E.V]=min (e.cap,a[u]); if(!VIS[E.V]) {VIS[E.V]=true; Q.push (E.V); } } } } if(D[t] = = INF)returnflow; Flow+=A[t]; Cost+ = a[t]*D[t]; for(inti = T; I! = S; i = edges[p[i]^1].v) {Edges[p[i]].cap-=A[t]; Edges[p[i]^1].cap + =A[t]; } } return-1;}voidinit () {memset (head,-1,sizeof(int)*vcnt); Edges.clear ();}intMain () {//freopen ("In.

The main topic: according to the order of the clockwise or counter-clockwise to give a multilateral point, to break this polygon into a n-2 triangle, the requirements of these triangular rows of the largest area of the Triangle area as small as possible, to find the minimum value.Ideas: with the interval DP can be easily solved, The polygon may be a concave shape, note that the triangle must be inside the polygon, so you can remove the split triangle contains other points, but other in the polyg

Title DescriptionDescriptionSchools need to connect n computers, and the cost of connection between the different 2 computers may be different. To save money, we consider the end of indirect data transfer, where one computer can indirectly connect to another computer via another computer.In order for any two computers to be connected (whether directly or indirectly), you need to use a network cable between several computers directly connected, now want to make the total cost of connection is the

The model layer of WPF is used to build data structuresFor these data structures, our general usage:A collection of this data structure type---> Fill data---> Bind foreground controlGood usage:The data structure of the model layer is raw material and is part of other data structures, we do not go directly to bind the foregroundCreate another data structure in the ViewModel layer, add the structure of the model layer here, the purpose of this new data structure is to be used to bind with the fore

Test instructionsEach edge in the figure has two weights (Cost,len), and a spanning tree is asked to minimize the sum (cost)/sum (len).Analysis:Transform to the maximum spanning tree + Newton iteration for the Edge s0*len-cost. S0 is a monotone iteration coefficient.Code:POJ 2728//sep9#include POJ 2728 Desert King parametric search solution optimal scale spanning tree

Every, villages is at different altitudes, they concluded, each channel between, villages needed a vertic Al Water lifter, which can lift water up or let water flow down. The length of the channel is the horizontal distance between the villages. The cost of the channel is the height of the lifter. You should notice the village is at a different altitude, and different channels can ' t share a lifter. Channels can intersect safely and no three villages is on the same line.As King David's prime s

http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1231Direct recursion.Save the maximum number of paths to the maximum value at the same time, and note the first column of the first row.Don't forget to take the mold.1#include 2#include 3#include 4#include 5 Const intMAXN =510;6 intDP[MAXN][MAXN],P[MAXN][MAXN],Q[MAXN][MAXN];7 intMain ()8 {9 intN;Ten while(~SCANF ("%d",N)) One { AMemset (DP,0,sizeof(DP)); - for(intI=1; i) { - for(intj=1; j) { thescanf"

It is important to carefully select the optimal MySQL Data Type. Why? Performance can be improved, and the principle is as follows: ● storage (memory, disk), thus saving I/O (retrieval of the same data) ● computing, and then reducing the CPU load data types can be divided into three types: number, character, and period (I) number ① classification ② it is meaningless to specify the width for the integer type. It is also probably to display the number o