Codeforces 786B Legacy (线段树优化建图)

Codeforces 786B Legacy (线段树优化建图)

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题意：\(n\)个点，有\(3\)种连边操作：1.将\(u\)指向\(v\)；2.将\(v\)指向编号在区间\([l,r]\)的点；3.将\([l,r]\)中的所有点指向\(v\)

做法：线段树优化建图。拓展一些新的节点来代表某些区间的点，然后，如果需要进行区间\([L,R]\)连边，那么就可以将\([L,R]\)拆成一些区间的组合，以此减少边的数目。于是可以利用线段树，将每个线段树上的节点新建出来，连接对应的区间即可，而因为如果只有一颗线段树不能对所有点跑最短路，所以我们建两棵线段树，一颗指向序列，另一颗从序列指出来。原本对应的区间边权设为0，跑最短路即可。

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#define IOS ios::sync_with_stdio(false)
#define pb push_back
#define Pii pair<int,int>
#define Ppi pair<Pii, int>
#define x first
#define y second
typedef long long ll;
const int N = 100020;
inline void read(int &x) {x = 0; int f = 1; char c = getchar();while(!isdigit(c)) {if(c=='-')f=-1;c=getchar();}while(isdigit(c)) {x=x*10+c-'0';c=getchar();}x *= f;
}
using namespace std;
bool ST;
int n, q, st;
struct edge {int e,w,nxt;} E[10000003];
int h[6000000], cc;
void add(int u, int v, int w) {E[cc].e = v; E[cc].w = w; E[cc].nxt = h[u]; h[u] = cc; ++cc;
}
int ID[2][N << 2], id;
void build(int p, int l, int r, int typ) {ID[typ][p] = ++id;if(!typ) for(int i = l; i <= r; ++i) add(ID[typ][p], i, 0);else for(int i = l; i <= r; ++i) add(i, ID[typ][p], 0);if(l == r) return;int mid = (l + r) >> 1;build(p<<1,l,mid,typ);build(p<<1|1,mid+1,r,typ);
}
void update(int p, int l, int r, int v, int L,int R, int w, int typ) {if(l == L && r == R) {if(!typ) add(v, ID[typ][p], w);else add(ID[typ][p], v, w);return;}int mid = (l + r) >> 1;if(R <= mid) update(p<<1,l,mid, v,L,R,w,typ);else if(L > mid) update(p<<1|1,mid+1,r, v,L,R,w,typ);else {update(p<<1,l,mid, v,L,mid,w,typ);update(p<<1|1,mid+1,r, v,mid+1,R,w,typ);}
}
struct node {int e; ll w;node(){}node(int a, ll b) {e=a;w=b;}bool operator < (const node &a) const {return w > a.w;}
};
ll dis[6000001];
int vis[6000001];
void dij(int st) {for(int i = 1; i <= id; ++i) dis[i] = __LONG_LONG_MAX__;priority_queue< node > q;q.push(node(st,0)); dis[st] = 0;while(!q.empty()) {node u = q.top(); q.pop();if(vis[u.e]) continue;vis[u.e] = 1;for(int i = h[u.e]; ~i ; i = E[i].nxt) {int v = E[i].e; ll w = E[i].w;if(dis[u.e] != __LONG_LONG_MAX__ && dis[v] > dis[u.e] + w) {dis[v] = dis[u.e] + w;q.push(node(v,dis[v]));}}}
}
int main() {read(n), read(q), read(st);memset(h, -1, sizeof(h));id = n;build(1,1,n,0); // tree -> []build(1,1,n,1); // [] -> treefor(int opt,u,v,w,l,r,i = 1; i <= q; ++i) {read(opt);if(opt == 1) read(u), read(v), read(w), add(u, v, w);else if(opt == 2) read(v), read(l), read(r), read(w), update(1,1,n, v,l,r,w, 0);else read(v), read(l), read(r), read(w), update(1,1,n, v,l,r,w, 1);}dij(st);for(int i = 1; i <= n; ++i) printf("%lld ",dis[i] == __LONG_LONG_MAX__ ? -1 : dis[i]); puts("");
}