For the first time the girls training analysis related questions:
The first is the first question:
How Many Tables
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include <algorithm>
5 #include <math.h>
6
7 using namespace std;
8
9 int pre[1001];
10 struct node{
11 int x;
12 int y;
13 }a[1001];
14
15 int find(int x){
16 int r=x;
17 while(pre[r]!=r)
18 r=pre[r];
19 return r;
20 }
21
22 int main(){
23 int N,M,i,num;
24 int T;
25 scanf("%d",&T);
26 while(T--){
27 scanf("%d%d",&N,&M);//N表示朋友个数,M表示关系条数
28 num = N; // N n between friends sitting up tables
29 for (i = 1; i <= N; i ++) // pre array initialization
30 pre [I] = I;
31 is for (I =. 1; I <= M; I ++) {
32 Scanf ( "% D% D ", a & [I] .x, & a [I] .y);
33 is here connected // friends know
34 is int Find B = (a [I] .x);
35 C int = Find (a [I ] .y);
36 IF (B = C) {!
37 [pre [C] = B;
38 is num--;
39}
40}
41 is the printf ( "% D \ n-", NUM); // for output formats have variable
42 is}
43 is return 0;
44 is}
So I decided to brush and check the water problem set, you can first learn she said Tuesday brush water problem, then gradually increase the difficulty, to build their confidence.
The following is a water theme A:
A - Smooth Traffic Project HDU 1232 (roughly meaning of the title)
A provincial survey of urban traffic, urban roads to get existing tables, the table lists the cities and towns each road directly connected. Target provincial government "Smooth Traffic Project" is to make the province between any two towns can implement traffic (but not necessarily directly connected to the road, as long as you can reach each other indirectly through road). Minimum asked how many roads also need to build?
Input
Test input contains several test cases. Each test case is given row of the first two positive integers, are the number of towns N (<1000) and road number M; M rows corresponding to the next M path, each row is given a pair of positive integers, respectively, the number two towns in direct communication path. For simplicity, the town numbered from 1 to N.
Note: You can communicate a number of roads between the two cities,
#Include. 1 <the iostream>
2 #include <stdio.h>
. 3
. 4 pre int [1001];
. 5
. 6 Find int (int X) {// find the root node
. 7 int X = R & lt;
. 8 the while (pre [R & lt]! = r) // return the root node
. 9 = R & lt pre [R & lt];
10 int X = I, J;
!. 11 the while (I = R & lt) {// path compressor
12 j = pre [i]; // change the upper before recording a temporary variable j at its value
13 pre [i] = r; // put his superiors directly to the root node, the ultimate Boss
14 I = j; // subsequently to change his original superior to its parent ultimate Boss
15}
16 return R & lt;
. 17}
18 is
. 19 the Join void (int a, int b) {
20 is // determines whether a and b are in communication
21 // If communication has been, you can not control
22 // If not connected, it the communication branch in which they are combined to
23 int fx = find (a) ;
Find int = FY 24 (B);
46 printf("%d\n",ans);
25 IF (! FX = FY)
26 pre [fx] = fy; // not optimized, the resulting tree may be deformed tree
27}
28
29 int main () {
30 int n-;
31 is the while (Scanf ( "% D", & n-) && n-= 0!) {
32 int m, I, a, B, ANS;
33 is for (I =. 1; I <= n-; I ++) // initialization pre array
34 is pre [I] = I;
35 // Memset (pre, 0 , pre + n-);
36 Scanf ( "% D", & m);
37 [for (I =. 1; I <= m; I ++) {
38 is Scanf ( "% D% D", & A, & B);
39 the Join ( A, B);
40}
41 is for (ANS = 0, I =. 1; I <= n-; I ++) {
42 is IF (pre [I] == I)
43 is ANS ++;
44 is}
45 ans- =. 1;
47}
return 0 48;
49}
Next came
B - ? Little hope maze | IT A Tree Is is HDU 1272 (roughly meaning of the questions)
Last Gardon maze castle little hope to play for a long time (see Problem B), and now she wanted to design a maze Gardon to let go. But she maze design ideas are not the same, she thought first of all of the channels should be two-way communication, that is if there is a passage room A and B, then both went to the room from the room A through B it can also be it went through the room a from the room B, to raise the difficulty, any small hope hope the two rooms has one and only one path can communicate (unless go back now). Little hope now her design to you, let you help her determine whether the design of her design ideas. For example the following example, the first two are eligible, but there are two ways to the final arrival 8 from 5.
Input
Input data comprises a plurality of sets, each set of data is an integer of 00 to the end of the list, indicates the number of two rooms a channel connection. No room is at least 1 and not more than 100,000. A blank line between each two sets of data.
Two end in the entire file 1.
In fact, I think this problem will start to see some faint, like training inside that title, but also need to look longer, in order to use the correct template. It seems my understanding of the disjoint-set is not enough. . . .
And this question was wrong several times, minor errors in the output of this you do not need to mention, is wrong on the order of some algorithms, the template is not cooked enough cooked ah!
The final AC Code:
#Include. 1 <the iostream>
2 #include <stdio.h>
. 3 the using namespace STD;
. 4
. 5 const int N = 100010;
. 6 pre int [N], S;
. 7 NUM int [N], V [N];
. 8
. 9 void the Init () {
10 for (int I =. 1; I <= N; I ++) {
. 11 pre [I] = I;
12 is V [I] = NUM [I] =. 1; // V [] recording state from is beginning. 1
13 is}
14}
15 Find int (int X) {// find the root node
16 int X = R & lt;
. 17 the while (! pre [R & lt] = R & lt) // return the root node
18 r = pre [r];
. 19 / int * X = I, J;
20 is the while (! I = R & lt) {// compression path
21 j = pre [i]; // j recorded at its value before the temporary variable to change the upper
22 pre [i ] = r; // put his superiors directly to the root node, the ultimate Boss
23 is I = J; // Next to change their original higher its superior ultimate Boss
} * 24/
25 return R & lt;
26 is}
27
28 int the Join (int a, int b) {
29 // determines whether a and b are in communication
30 FX int = Find (a);
31 is int FY = Find (b);
32 if (fx == fy) {// if communication at this time so that it will produce the tree circuit
33 is return. 1;
34 is}
35 the else {
36 pre [FY] = FX; // do not belong to the same two rooms as a connection tree
37 [NUM [FX] + = NUM [FY];
38 is S = NUM [FX];
39 return 0;
40 // not optimized, the resulting tree may be deformed tree
41 is}
42 is}
43 is
44 is int main () {
45 int I, A, B;
46 is the while (Scanf ( "% D% D", & A, & B) && (A = -. 1 || B = -. 1)!!) {
47 the Init ();
48 IF ( a == 0 && b == 0) {// Analyzing empty tree where
49 printf("Yes\n");
50 continue;
51 }
52 int ok=join(a,b),n=v[a]+v[b]; //ok记录是否存在环，n记录节点总数
53 v[a]=v[b]=0;
54 for(i=2;i<=N;i++){
55 scanf("%d%d",&a,&b);
56 if(a==0&&b==0)
57 break;
58 n+=(v[a]+v[b]);
59 v[a]=v[b]=0;
60 ok+=join(a,b);
61 }
62 if(ok)
63 printf("No\n");
64 else{
65 if(s==n)
66 printf("Yes\n");
67 else
68 printf("No\n");
69 }
70 }
71 return 0;
72 }
For summary disjoint-set of the first here, but also greatly enhance the learning efforts.