ccpc

链接:https://www.zhixincode.com/contest/24/problem/E?problem_id=353
思路:学习了线段树优化建图的方法,现在总结一下单树的优化。单树适用于像本题这样拓扑排序,不同于最短路,如果没有限制条件的话父区间是不需要向子区间走的(如果是最短路既需要上又需要下寻找路径,所以需要两棵树)。所以本题只用一棵向上连边的线段树即可。对于某段区间到某个点,我们只需要找到那段区间对应的结点,连向限制点的节点即可。然后我们从所有入度为0的叶节点开始走,用优先队列做一遍拓扑排序即可。

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using namespace std;
#define maxn 100010
#define ll long long


int n;
struct {
int from, to, dist;
};
vector<int> G[maxn<<3];
vector<edge> edges;
int tot;
int ls[maxn<<3], rs[maxn<<3];
int _o;
int t1,t2;
ll deg[maxn<<3];
vector<ll> res;
bool vis[maxn<<3];
ll val[maxn<<3];
const ll inf = 1e18;
int pos[maxn];

void init(){
for(int i = 0; i <= 2 * n; i++)G[i].clear();
edges.clear();
}

void addedge(int from, int to, int dist){
edges.push_back(edge{from, to, dist});
tot = edges.size();
G[from].push_back(tot - 1);
deg[to]++;
}

struct node {
ll v;
int id;
bool operator<(const node &r)const{
return v > r.v;
}
}b[maxn];

void build(int &o, int l, int r){
o = ++_o;
if(l < r) {
int mid = l + r >> 1;
build(ls[o], l, mid);
build(rs[o], mid + 1, r);
addedge(ls[o], o, 0), addedge(rs[o], o, 0);
return;
}
val[o] = b[l].v;
pos[l] = o;
}

void update(int o, int tl, int tr, int l, int r, int x){
if(tr < l || r < tl)return;
if(l <= tl && r >= tr){
addedge(o, x, 0);//区间节点向叶节点连边
return;
}
int mid = tl + tr >> 1;
update(ls[o], tl, mid, l, r, x);
update(rs[o], mid + 1, tr, l, r, x);
}

priority_queue<node> q;

//找到所有入度为0的叶节点
void insert(int o, int l, int r){
if(!deg[o]) {
q.push(node{val[o], o});
}
if(l == r)return;
int mid = l + r >> 1;
insert(ls[o], l, mid);
insert(rs[o], mid + 1, r);
}

int main(){
ios::sync_with_stdio(false);
cin >> n;
init();
for(int i = 1; i <= n; i++){
cin >> b[i].v;
b[i].id = i;
}
build(t1, 1, n);
for(int i = 1; i <= n; i++){
int idd = b[i].id;
int u = b[i].v % n;
u++;
if(u == idd)continue;
if(u < idd){
update(t1, 1, n, u, idd - 1, pos[idd]);
}
else{
update(t1, 1, n, 1, idd - 1, pos[idd]);
update(t1, 1, n, u, n, pos[idd]);
}
}
//做一遍拓扑排序即可
insert(t1, 1, n);
while(!q.empty()){
node u = q.top();
q.pop();
if(vis[u.id])continue;
vis[u.id] = true;
if(u.v)
res.push_back(u.v);
for(int i = 0; i < G[u.id].size(); i++){
edge &e = edges[G[u.id][i]];
int v = e.to;
if(vis[v])continue;
deg[v]--;
if(!deg[v])q.push(node{val[v], v});
}
}
for(int i = 0; i < res.size(); i++){
cout << res[i] << (i == res.size() - 1 ? 'n' : ' ');
}
return 0;
}