Codeforces 786B. Legacy 线段树+spfa

题目大意:

给定一个(n)的点的图。求(s)到所有点的最短路
边的给定方式有三种:

  • (u o v)
  • (u o [l,r])
  • ([l,r] o v)
    (q)为给定边的次数,有(n,q leq 10^5)

题解

类比于线段树优化网络流建图
写一个线段树优化最短路建图即可。

#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
    x=0;char ch;bool flag = false;
    while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
    while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
#define rg register int
#define rep(i,a,b) for(rg i=(a);i<=(b);++i)
#define per(i,a,b) for(rg i=(a);i>=(b);--i)
const int maxn = 100010;
struct Edge{
    int to,dis;
    Edge(){}
    Edge(const int &a,const int &b){
		to = a;dis = b;
    }
};
vector<Edge>ve[maxn*10];
inline void add(int u,int v,int d){
    ve[u].push_back(Edge(v,d));
}
int T1[maxn<<2],T2[maxn<<2],nodecnt;
void build1(int rt,int l,int r){
    T1[rt] = ++ nodecnt;
    if(l == r){
		add(l,T1[rt],0);
		return ;
    }
    int mid = l+r >> 1;
    build1(rt<<1,l,mid);
    build1(rt<<1|1,mid+1,r);
    add(T1[rt<<1],T1[rt],0);
    add(T1[rt<<1|1],T1[rt],0);
}
void build2(int rt,int l,int r){
    T2[rt] = ++ nodecnt;
    if(l == r){
		add(T2[rt],l,0);
		return ;
    }
    int mid = l+r >> 1;
    build2(rt<<1,l,mid);
    build2(rt<<1|1,mid+1,r);
    add(T2[rt],T2[rt<<1],0);
    add(T2[rt],T2[rt<<1|1],0);
}

int p,val;
void query1(int rt,int l,int r,int L,int R){
    if(L <= l && r <= R){
		add(T1[rt],p,val);
		return ;
    }
    int mid = l+r >> 1;
    if(L <= mid) query1(rt<<1,l,mid,L,R);
    if(R >  mid) query1(rt<<1|1,mid+1,r,L,R);
}
void query2(int rt,int l,int r,int L,int R){
    if(L <= l && r <= R){
		add(p,T2[rt],val);
		return ;
    }
    int mid = l+r >> 1;
    if(L <= mid) query2(rt<<1,l,mid,L,R);
    if(R >  mid) query2(rt<<1|1,mid+1,r,L,R);
}
ll dis[maxn*10];bool inq[maxn*10];
void spfa(int s){
    memset(dis,0x3f,sizeof dis);
    dis[s] = 0;inq[s] = true;
    queue<int>q;q.push(s);
    while(!q.empty()){
		int u = q.front();q.pop();
		for(vector<Edge>::iterator it = ve[u].begin();it != ve[u].end();++ it){
		    if(dis[it->to] > dis[u] + it->dis){
				dis[it->to] = dis[u] + it->dis;
				if(!inq[it->to]){
				    inq[it->to] = true;
				    q.push(it->to);
				}
		    }
		}inq[u] = false;
    }
}
int main(){
    int n,q,s;read(n);read(q);read(s);
    int t,u,v,l,r;
    nodecnt = n;
    build1(1,1,n);build2(1,1,n);
    while(q--){
		read(t);
		if(t == 1){
		    read(u);read(v);read(val);
		    add(u,v,val);
		}else if(t == 2){
		    read(p);read(l);read(r);read(val);
		    query2(1,1,n,l,r);
		}else if(t == 3){
		    read(p);read(l);read(r);read(val);
		    query1(1,1,n,l,r);
		}
    }
    spfa(s);
    rep(i,1,n){
		if(dis[i] != dis[0]) printf("%lld",dis[i]);
		else printf("%d",-1);
		if(i != n) putchar(' ');
		else putchar('
');
    }
    return 0;
}
原文地址:https://www.cnblogs.com/Skyminer/p/6952386.html