解題思路：這道題并查集很容易，合并時(shí)找到父節(jié)點(diǎn)就直接加上去就ok了。關(guān)鍵是如何求K大數(shù)，我一直在想用線段樹怎么寫，一開始想如果直接記錄數(shù)的大小那肯定是沒戲了，借鑒了一下別人的思路：區(qū)間[a,b]記錄的是所有的數(shù)里面，等于a，a+1，a+2，......，b-1，b的個(gè)數(shù)?？吹竭@里就應(yīng)該明白了，這里線段樹的用法是把它看做是一個(gè)1-n的數(shù)軸。到時(shí)候要修改某一個(gè)數(shù)，就直接在數(shù)軸上修改它。至于記錄[a,b]的個(gè)數(shù)，是為了找到K大數(shù)。

參考博客：http://blog.sina.com.cn/s/blog_6fd8e0fe0100v89n.html

#include<iostream> #include<cstdio> #include<cstring> using namespace std;const int maxn = 200005; struct Segment {int l,r;int sum; }tree[maxn<<2]; int n,m,fa[maxn],tot[maxn];int find(int x) {if(fa[x] == x) return x;return fa[x] = find(fa[x]); }void build(int rt,int l,int r) {tree[rt].l = l, tree[rt].r = r;if(tree[rt].l == 1) tree[rt].sum = n;else tree[rt].sum = 0;if(l == r) return;int mid = (l + r) >> 1;build(rt<<1,l,mid);build(rt<<1|1,mid+1,r); }void update(int rt,int pos,int val) {tree[rt].sum += val;if(tree[rt].l == tree[rt].r) return;int mid = (tree[rt].l + tree[rt].r) >> 1;if(pos <= mid) update(rt<<1,pos,val);else update(rt<<1|1,pos,val); }int query(int rt,int k) {if(tree[rt].l == tree[rt].r) return tree[rt].l;int mid = (tree[rt].l + tree[rt].r) >> 1;if(tree[rt<<1|1].sum >= k) return query(rt<<1|1,k);else return query(rt<<1,k - tree[rt<<1|1].sum); }int main() {int op,a,b,x,y;while(scanf("%d%d",&n,&m)!=EOF){for(int i = 1; i <= n; i++)fa[i] = i, tot[i] = 1;build(1,1,n);for(int i = 1; i <= m; i++){scanf("%d",&op);if(op){scanf("%d",&a);printf("%d\n",query(1,a));}else{scanf("%d%d",&a,&b);x = find(a);y = find(b);if(x == y) continue;fa[y] = x;update(1,tot[x],-1);update(1,tot[y],-1);update(1,tot[x]+tot[y],1);tot[x] += tot[y];tot[y] = 0;}}}return 0; }

用樹狀數(shù)組求k大數(shù)：

#include<iostream> #include<cstdio> #include<cstring> using namespace std;const int maxn = 200005; struct Tree {int n,c[maxn];void init(int n){this->n = n;memset(c,0,sizeof(c));}int lowbit(int x){return x & -x;}void update(int x,int d){while(x <= n){c[x] += d;x += lowbit(x);}}int sum(int x){int ans = 0;while(x > 0){ans += c[x];x -= lowbit(x);}return ans;} }tree; int n,m,fa[maxn],cnt[maxn];int find(int x) {if(fa[x] == x) return x;return fa[x] = find(fa[x]); }int main() {int op;while(scanf("%d%d",&n,&m)!=EOF){tree.init(n);tree.update(1,n);for(int i = 1; i <= n; i++) fa[i] = i,cnt[i] = 1;while(m--){scanf("%d",&op);if(op == 0){int i,j;scanf("%d%d",&i,&j);int f1 = find(i);int f2 = find(j);if(f1 != f2){tree.update(cnt[f1],-1);tree.update(cnt[f2],-1);fa[f2] = f1;cnt[f1] += cnt[f2];tree.update(cnt[f1],1);}}else{int k,l = 1,r = n,mid,ans;scanf("%d",&k);while(l <= r){ mid = (l + r) >> 1;int tmp = tree.sum(n) - tree.sum(mid-1);if(tmp >= k){ans = mid;l = mid + 1;}else r = mid - 1;}printf("%d\n",ans);}}}return 0; }

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