題干：

Mr Wang wants some boys to help him with a project. Because the project is rather complex,?the more boys come, the better it will be. Of course there are certain requirements.?

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.?
InputThe first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)OutputThe output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.?
Sample Input 4 1 2 3 4 5 6 1 6 4 1 2 3 4 5 6 7 8Sample Output 4 2 Hint A and B are friends(direct or indirect), B and C are friends(direct or indirect), then A and C are also friends(indirect).In the first sample {1,2,5,6} is the result. In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.

解題報告：????

????裸一個并查集，最后查的時候記錄一下最大值就好了。

ac代碼：

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> const int MAX = 10000000 + 5; using namespace std;int f[MAX]; int sum[MAX]; int getf(int v) {return f[v]==v? v : f[v]=getf( f[v] ) ; } void merge(int u,int v) {int t1,t2;t1=getf(u);t2=getf(v);if(t1!=t2) {f[t2]=t1;}} void init() {for(int i = 0; i<MAX; i++) {f[i]=i;sum[i]=0;} } int maxx,u,v; int main() {int n;while(~scanf("%d",&n)) {if(n==0) {printf("1\n");continue; }init();maxx=0;while(n--) {scanf("%d %d",&u,&v);merge(u,v);if(u>maxx||v>maxx) maxx=max(u,v);}for(int i = 1; i<=maxx; i++) {sum[getf(f[i])] ++;}printf("%d\n",*max_element(sum+1,sum+maxx+1));}return 0 ;}

ac代碼2：（空間優化了一點點，只開了一個數組）

#include<cstdio> #include<algorithm> #include<cstring> using namespace std;int pre[10000005]; int fin(int x) {if(pre[x]==x){return x;}else{return pre[x]=fin(pre[x]);} }void join(int x,int y) {int t1=fin(x);int t2=fin(y);if(t1!=t2){pre[t1]=t2;} }int main() {int n;while(~scanf("%d",&n)){if(n==0)//特判，被坑在這了{printf("1\n");continue;}int maxn=0,a,b;for(int i=0;i<=10000004;i++){pre[i]=i;}for(int i=0;i<n;i++){scanf("%d%d",&a,&b);join(a,b);if(max(a,b)>maxn)//優化技巧{maxn=max(a,b);}}int sum=0,sum1=0;for(int i=1;i<=maxn;i++){sum1=0;if(pre[i]==i)for(int j=1;j<=maxn;j++){if(i==fin(j)){sum1++;}}sum=max(sum,sum1);//更細最大值}printf("%d\n",sum);}return 0; }

總結：

????n==0時別忘特殊處理一下！！！

????有個地方時間優化的十分正確 ?maxx的記錄。

????還可以再時間優化：merge的時候，在合并點的同時（t2合并入t1），sum[t1]+=sum[t2];(即把t2的成員也都加入t1)

????有個缺點 占內存太多啦！有沒有更好的方法？

總結

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