Pick up Zoj a week to solve problems

ZOJ 2734 Exchange Cards Main topic:

Given a value of N, and a pile of cards, each card has a value of values and a number of numbers. Use any card to form N.

For example n = ten, Cards (10,2) (7,2) (5,3) (2,2) (1,5), then 10 = 10,10 =7+3+1,10=5+5 ...

Thinking Analysis:

Since done before 1204, know that the problem is naked search, directly with the DFS violence can be obtained.

Can do optimization processing is-this process is greedy greedy, first from the big to small to take over, so, you can do a step descending sort of preprocessing.

As in the above example: I chose 7, then I would not choose 10, because plus 10 is too big, beyond the N.

Summary of Knowledge points:

1) DFS walk: Go forward along the feasible solution space;

DFS (index, number, sum, target);

2) Cycle Away method: Select the next starting point enumeration;

DFS (++index, 1, sum, target);

Code:

#include <iostream>#include<vector>#include<map>#include<algorithm>#include<fstream>using namespacestd;structcards{intnum; intvalue; BOOL operator< (Constcards& card)Const    {        returnValue >Card.value; }};intTotalcardscount;vector<Cards>inputvalues;voidDfsintIndexintNumberintSuminttarget) {    if(sum = = target)//Summary of Records{Totalcardscount++; return; }    if(Index = = inputvalues.size ())return;//length//can add, deep ing ...    if(sum + inputvalues[index].value <= target && number <=inputvalues[index].num) {Sum+=Inputvalues[index].value; number++;        DFS (index, number, sum, target); number--; Sum-=Inputvalues[index].value; } DFS (++index,1, sum, target);}intMain () {//ifstream cin ("2734.txt");    intT, I, Target; intc =0; intN; BOOLFirst =true;  while(Cin >>target) {        if(!First ) {cout<<Endl; } First=false; C=1; CIN>>N;        Inputvalues.clear (); inttmp;  for(i =1; I <= N; i++) {Cards cd; CIN>>Cd.value; CIN>>Cd.num;        Inputvalues.push_back (CD);        } std::sort (Inputvalues.begin (), Inputvalues.end ()); Totalcardscount=0; DFS (0,1,0, Target); cout<< Totalcardscount <<Endl; }    return 0;}

ZOJ 1947 the Settlers of Catan Main topic:

An--the graph is given to find the longest path of the non-graph without permission longest path. That is, find a path in the graph where the edges are not repeated (points can be repeated), and the number of edges of the resulting path is the desired one.

such as: 3 points, two edges, (0,1), the longest path is 2.

The maximum size of the graph is 25*25.

Thinking Analysis:

See this topic, try to go online search for the longest path of knowledge points, but always find no useful. Finally, I put the starting point on the concept of the Euler loop, and then I understand the Hamiltonian loop, and come to a conclusion that the longest path is more extensive than the Euler circuit.

Accidentally browsing to Wikipedia, knowing that this is a NP problem, its algorithm can only be a brute force enumeration. Better understand that if it is the longest path to the graph, then the topological sort preprocessing should be done first.

As a result, DFS searches are performed directly on each point, saving the longest path at a time.

Summary of Knowledge points:

1) The Euler circuit: a loop that passes all the edges once and all the vertices one at a time.

2) Hamiltonian circuit: a loop that passes through all vertices once in the diagram.

Code:

#include <stdio.h>#include<iostream>#include<vector>#include<map>#include<Set>#include<algorithm>#include<cstring>#include<fstream>#include<list>using namespacestd;intgraph[ -][ -];BOOLvisited[ -][ -];intThelongestpath;voidDfsintRootintd) {    if(D >Thelongestpath) {Thelongestpath=D; }     for(inti =0; I < -; i++)    {        if(Root = = i)Continue; if(Graph[root][i] = =1&& Visited[root][i] = =0) {Visited[root][i]= Visited[i][root] =1; ++D;            DFS (i, d); --D; Visited[root][i]= Visited[i][root] =0; }    }}intmain1947 () {//fstream cin ("1947.txt");    intN, M;  while(Cin >> N >>m) {if(n = =0&& m = =0) Break; memset (graph,0,sizeof(graph)); intstart, end;  for(inti =0; I < m; i++) {cin>> Start >>end; Graph[start][end]= Graph[end][start] =1; } Thelongestpath=0;  for(inti =0; I < n; i++) {memset (visited,0,sizeof(visited)); DFS (i,0); } cout<< Thelongestpath <<Endl; }    return 0;}

ZOJ 1978 Assistant Required Main topic:

Given a queue, 2, 3, ... N, each time take the first element of the team as a lucky element, and then the subsequent every I to drag out to work. For example, the first time 2 is lucky, 4,6,8 ... The second time 3 is lucky, 9,15,21 ... Going to work ...

Ask for the K lucky number.

Thinking Analysis:

From 2 wrote the sequence analysis of 20, found that the prime number is similar, and even made a prime table.

But when testing the 20th lucky number, the Sample out of the given is 83, I hit 71 (or 73, specifically forget), found wrong, and then analysis, found and the prime table a little difference.

Prime table: A multiple of I is removed each time;

Lucky numbers (for the moment) table: Every time I remove the number.

Modify it, you can.

Summary of Knowledge points:

1) prime list;

Code:

#include <iostream>#include<vector>#include<map>#include<string>#include<algorithm>#include<fstream>using namespacestd;Const intMax_prime =3001;Const intSearch_int =34000;intPrime_like[max_prime];BOOL  is[Search_int];voidmake_prim_table () { for(inti =0; i < max_prime; i++)    {         is[I] =0; }    intnum =1;  for(inti =2; i < Search_int; i++)    {        if( is[I] = =1)Continue; if( is[I] = =0) {Prime_like[num++] =i; if(num = = max_prime) Break; }        if(i = =2)        {             for(intK = i; K < Search_int; K + =i) { is[K] =1; }        }        Else        {            intNumber =0;  for(intK = i +1; K < Search_int; k++)            {                if( is[K] = =0)//already out of the team                {                    if(++number = =i) { is[K] =1; number=0; }                }            }        }    }}intmain1978 () {make_prim_table (); //fstream cin ("1978.txt");    intN;  while(true) {cin>>N; if(n = =0) Break; cout<< Prime_like[n] <<Endl; }    return 0;}

Acceptted is the best thing for you ~

Pick up Zoj a week to solve problems