CF786B Legacy

Posted 2021-02-09 lm-lbg

好久不写题解了QAQ

传送门：https://www.luogu.org/problemnew/show/CF786B

很巧妙的一道题

考虑建两颗线段树，一颗out,维护出去的边，一颗in,维护进来的边

这样的话，所有的加进来的边，都是从out连向in,而且最后的最短路，实际上求得是out中的点到in中的点的最短路（虽然代码上体现不出）

in,out中的连边方式，也是考虑上述结论得出的

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;
#define ll long long 
inline int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < ‘0‘ || ch > ‘9‘)
    {
        if(ch == ‘-‘) op = -1;
        ch = getchar();
    }
    while(ch >= ‘0‘ && ch <= ‘9‘)
    {
         (ans *= 10) += ch - ‘0‘;
         ch = getchar();
    }
    return ans * op;
}
const int maxn = 2e5 + 5;
struct node
{
    int to,cost;
};
vector<node> e[maxn * 10];
int cnt;
struct segement_tree
{
    int in[maxn << 2],out[maxn << 2];
    #define ls i << 1
    #define rs i << 1 | 1
    void build(int i,int l,int r)
    {
        if(l == r)
        {
            in[i] = l,out[i] = l;
            return;
        }
        int mid = l + r >> 1;
        build(ls,l,mid),build(rs,mid + 1,r);
        in[i] = ++cnt,out[i] = ++cnt;
        e[in[i]].push_back((node){in[ls],0});
        e[in[i]].push_back((node){in[rs],0});
        e[out[ls]].push_back((node){out[i],0});
        e[out[rs]].push_back((node){out[i],0});
    }
    void addein(int i,int l,int r,int ql,int qr,int u,int w)
    {
        if(l == ql && r == qr) 
        {
            e[u].push_back((node){in[i],w});
            return;
        }
        int mid = l + r >> 1;
        if(qr <= mid) addein(ls,l,mid,ql,qr,u,w);
        else if(ql > mid) addein(rs,mid + 1,r,ql,qr,u,w);
        else addein(ls,l,mid,ql,mid,u,w),addein(rs,mid + 1,r,mid + 1,qr,u,w);
    }
    void addeout(int i,int l,int r,int ql,int qr,int u,int w)
    {
        if(l == ql && r == qr)
        {
            e[out[i]].push_back((node){u,w});
            return;
        }
        int mid = l + r >> 1;
        if(qr <= mid) addeout(ls,l,mid,ql,qr,u,w);
        else if(ql > mid) addeout(rs,mid + 1,r,ql,qr,u,w);
        else addeout(ls,l,mid,ql,mid,u,w),addeout(rs,mid + 1,r,mid + 1,qr,u,w);
    }
}T;
struct pnode
{
    ll id,dis;
    bool operator < (const pnode& a) const
    {
        return dis > a.dis;
    }
};
ll dis[maxn * 10];
priority_queue<pnode> p;
int s;
void dij()
{
    memset(dis,0x3f,sizeof(dis));
    pnode x;
    x.dis = 0,x.id = s;
    dis[s] = 0;
    p.push(x);
    while(p.size())
    {
        pnode x = p.top();
        p.pop();
        int u = x.id;
        if(dis[u] != x.dis) continue;
        for(int cur = 0;cur < e[u].size();cur++)
        {
            int v = e[u][cur].to,w = e[u][cur].cost;
            if(dis[v] > dis[u] + w)
            {
                dis[v] = dis[u] + w;
                pnode ne;
                ne.id = v,ne.dis = dis[v];
                p.push(ne); 
            }
        }
    }
}
int main()
{
    int n = read(),q = read();
    s = read();
    cnt = n; 
    T.build(1,1,n);
    for(int i = 1;i <= q;i++)
    {
        int op = read(); 
        if(op == 1) {int u = read(),v = read(),w = read(); e[u].push_back((node){v,w}); continue;}
        int u = read(),l = read(),r = read(),w = read(); 
        if(op == 2) T.addein(1,1,n,l,r,u,w);
        if(op == 3) T.addeout(1,1,n,l,r,u,w);
    }
    dij();
    for(int i = 1;i <= n;i++) if(dis[i] == 0x3f3f3f3f3f3f3f3f) printf("-1 "); else printf("%lld ",dis[i]);
}