并查集基础题 POJ1611 POJ1703 POJ1988

刚学习了并查集,于是找了几道并查集的基础题来做,感觉掌握的不是很牢固,POJ1182食物链和ZOJ3261星星能量这两题都没能做出来..先放着,以后再做.

 

 

POJ1611 确定SARS病毒的传播圈,并查集的第一题,因为在学习阶段,所以就照着两种写法都写了一遍

1.初始化每个元素为独立集合,用rank来标记每个集合的人数,初始为1,然后每次合并跟新根的rank值,并且都是将较小的节点作为根,最后查看rank[0]就可以了

#include <iostream>#include <stdio.h>using namespace std;struct{int rank;int parent;}ar[30010];void init(int n){for(int i=0;i<n;i++){ar[i].parent=i;ar[i].rank=1;}}int find(int x){if(x!=ar[x].parent){ar[x].parent=find(ar[x].parent);}return ar[x].parent;}void merge(int x,int y){int px=find(x);int py=find(y);if(px==py)return;if(px<py){ar[py].parent=px;ar[px].rank+=ar[py].rank;}else{ar[px].parent=py;ar[py].rank+=ar[px].rank; }}int main(){while(1){int n,m,group;scanf("%d%d",&n,&m);if(!n&&!m)break;init(n);for(int i=0;i<m;i++){scanf("%d",&group);int a,b;scanf("%d",&a);for(int j=1;j<group;j++){scanf("%d",&b);merge(a,b);}}printf("%d/n",ar[0].rank);}}  

 

2.parent初始化为-1,每次合并的时候将parent值相加,最后parent值的相反数就是这个集合的人数.这种方法就是在回溯缩短路径的时候要注意返回的跟并不是根的parent,而是根的值.

#include <iostream>#include <cstdio>using namespace std;int pa[30010];void make_set(int n){for(int i=0;i<n;i++){pa[i]=-1; }}int find_set(int x){int p=x;while(pa[p]>0)p=pa[p];while(x!=p){int t=pa[x];pa[x]=p;x=t;}return x;}void union_set(int x,int y){x=find_set(x);y=find_set(y);if(x!=y){if(pa[x]<pa[y]){pa[x]+=pa[y];pa[y]=x;}else{pa[y]+=pa[x];pa[x]=y;}}}int main(){while(1){int n,m,k;scanf("%d%d",&n,&m);if(!(n||m))break;make_set(n); while(m-->0){scanf("%d",&k);int a,b;scanf("%d",&a);k--;while(k-->0){scanf("%d",&b);union_set(a,b);} }printf("%d/n",0-pa[find_set(0)]); }}  

 

两种方法的时间一样,后一种方法的占用内存稍微少一点

 

 

POJ1703 分两派黑帮,是食物链的简化版,想了一会毫无思路,只能靠度娘了,哎,什么时候才能不再是初学者啊~~

两种方法,都挺好的,都学习一下吧.

 

1.设置一个opp数组,x的敌人为opp[x], 分情况合并集合

#include <iostream> #include <cstdio>using namespace std;int opp[100010];int parent[100010];void make_set(int n){for(int i=1;i<=n;i++){parent[i]=i;opp[i]=0; }}int find_set(int x){if(x!=parent[x]){parent[x]=find_set(parent[x]);}return parent[x];}void union_set(int x,int y){int fx=find_set(x);int fy=find_set(y);if(fx==fy)return;parent[fy]=fx;}int main(){int nCase;scanf("%d",&nCase);while(nCase-->0){int n,m,a,b;char c;scanf("%d%d",&n,&m);make_set(n);while(m-->0){getchar();//注意吸收回车 scanf("%c%d%d",&c,&a,&b);if(c=='D'){//分情况讨论 if(opp[a]==0&&opp[b]==0){opp[a]=b;opp[b]=a;}else if(opp[a]==0){opp[a]=b;union_set(a,opp[b]);}else if(opp[b]==0){opp[b]=a;union_set(b,opp[a]);}else{union_set(a,opp[b]);union_set(b,opp[a]);}}else{int x=find_set(a);int y=find_set(b);int z=find_set(opp[b]);if(x==y){//a和b在相同的帮派 printf("In the same gang./n");}else if(x==z){//a和b的对手在同一个集合,说明是不同帮派 printf("In different gangs./n");}else{printf("Not sure yet./n");}}}}return 0;} 

 

2.用rank标记该点和根的关系,相同帮派为0,不同帮派为1 感觉重点就是回溯跟新rank.

#include <iostream> #include <cstdio>using namespace std;int rank[100010];int parent[100010];void make_set(int n){for(int i=0;i<n;i++){parent[i]=i;rank[i]=0;//0代表和父节点同类,1代表异类 } }int find_set(int x){int t=parent[x];if(x!=parent[x]){parent[x]=find_set(t);rank[x]=(rank[t]+rank[x])%2;//回溯跟新 和父节点相同为0,不同为1 }return parent[x];}void union_set(int x,int y){int fx=find_set(x);int fy=find_set(y);if(fx==fy)return;parent[fx]=fy;//if(!rank[x]&&rank[y])rank[fx]=0;//if(rank[x]&&!rank[y])rank[fx]=0;//if(rank[x]&&rank[y])rank[fx]=1;//if(!rank[x]&&!rank[y])rank[fx]=1;rank[fx]=(rank[x]+rank[y]+1)%2;//根据x与其根,y与其根,x与y的关系,推导出fx与fy的关系 }int main(){int nCase;scanf("%d",&nCase);while(nCase-->0){int n,m,a,b;char c;scanf("%d%d",&n,&m);make_set(n);while(m-->0){getchar();scanf("%c%d%d",&c,&a,&b);if(c=='D'){union_set(a,b); }else{int x=find_set(a);int y=find_set(b);if(x==y){if(rank[a]==rank[b]){printf("In the same gang./n");}else printf("In different gangs./n");}else{printf("Not sure yet./n");}}}}} 

 

 

POJ1988 每次移动a和b的所在列,最后查询c下面的方块数

其中below和stack两个参数一个是某个方块下面的方块数 和 某一列的高度,感觉这两个参数十分巧妙,注意在路径压缩时以及合并时要及时跟新,另外查询时也要进行一次跟新.

/*Cube StackingTime Limit: 2000MSMemory Limit: 30000KTotal Submissions: 11813Accepted: 3946Case Time Limit: 1000MSDescriptionFarmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations: moves and counts. * In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y. * In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value. Write a program that can verify the results of the game. Input* Line 1: A single integer, P * Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X. Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself. OutputPrint the output from each of the count operations in the same order as the input file. Sample Input6M 1 6C 1M 2 4M 2 6C 3C 4Sample Output102*/#include <cstdio>using namespace std;int parent[30010];int stack[30010]; //这一堆的总数量 int below[30010]; //每个方块到它父节点之间的方块数 int n;void make_set(){for(int i=0;i<30010;i++){parent[i]=i; below[i]=0;stack[i]=1;}}int find_set(int x){int temp=parent[x];if(parent[x]!=x){parent[x]=find_set(parent[x]);below[x]+=below[temp];//回溯压缩路径,同时要跟新below参数}return parent[x];}void union_set(int x,int y){int px=find_set(x);int py=find_set(y);parent[px]=py;//x移到y上,即x堆并入y堆 below[px]+=stack[py]; //x到根的距离即为y的高度 stack[py]+=stack[px];//堆的总数量跟新 }int main() {scanf("%d",&n);make_set();for(int i=0;i<n;i++){char c;int a,b; scanf("/n%c",&c);if(c=='M'){scanf("%d%d",&a,&b);union_set(a,b);}else{scanf("%d",&a);find_set(a);//跟新a的参数 printf("%d/n",below[a]); }}return 0;} 

 

 

做了几题,感觉掌握了并查集的方法,但还没有掌握并查集的思想,任重而道远,继续努力吧!!!