Legacy 【CodeForces - 786B】【线段树优化建图】

题目链接CF-786B

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题意：有N个点，Q次操作，图的起点是S，问经过Q次操作之后，S到每个点的最短路。

1. 增加从点v到点u的距离w的边
2. 增加从点v到点[l, r]区间的每个点的距离w的边
3. 增加从区间[l, r]到点v的距离为w的边

  于是，我们可以开两棵线段树，分别表示出入的情况，然后操作结束之后跑一次堆优化的Dijkstra即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define BIG_INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
int N, Q, S, tot, rt1, rt2, head[maxN << 3], cnt;
struct Eddge
{
    int nex, to; ll val;
    Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), val(c) {}
} edge[maxN << 5];
inline void addEddge(int u, int v, ll w)
{
    edge[cnt] = Eddge(head[u], v, w);
    head[u] = cnt++;
}
namespace Segement
{
    inline int new_node() { ++tot; head[tot] = -1; return tot; }
    int lc[maxN << 3], rc[maxN << 3];
    void build(int &u, int fa, int l, int r, int op)
    {
        u = new_node();
        if(fa)
        {
            if(op) addEddge(u, fa, 0);
            else addEddge(fa, u, 0);
        }
        if(l == r) return;
        int mid = HalF;
        build(lc[u], u, l, mid, op);
        build(rc[u], u, mid + 1, r, op);
    }
    inline int point_query(int rt, int l, int r, int qx)
    {
        if(l == r) return rt;
        int mid = HalF;
        if(qx <= mid) return point_query(lc[rt], l, mid, qx);
        else return point_query(rc[rt], mid + 1, r, qx);
    }
    inline void Range_query(int rt, int l, int r, int ql, int qr, int u, int w, int op)
    {
        if(ql <= l && qr >= r)
        {
            if(op) addEddge(u, rt, w);
            else addEddge(rt, u, w);
            return;
        }
        int mid = HalF;
        if(qr <= mid) Range_query(lc[rt], l, mid, ql, qr, u, w, op);
        else if(ql > mid) Range_query(rc[rt], mid + 1, r, ql, qr, u, w, op);
        else { Range_query(lc[rt], l, mid, ql, qr, u, w, op); Range_query(rc[rt], mid + 1, r, ql, qr, u, w, op); }
    }
};
using namespace Segement;
namespace Dijkstra
{
    ll dis[maxN << 3];
    struct node
    {
        int id; ll val;
        node(int a=0, ll b=0):id(a), val(b) {}
        friend bool operator < (node e1, node e2) { return e1.val > e2.val; }
    };
    priority_queue<node> Que;
    void Dij()
    {
        for(int i=1; i<=tot; i++) dis[i] = BIG_INF;
        int tmp = point_query(rt1, 1, N, S);
        dis[tmp] = 0;
        Que.push(node(tmp, 0));
        while(!Que.empty())
        {
            node now = Que.top(); Que.pop();
            int u = now.id;
            if(dis[u] < now.val) continue;
            for(int i=head[u], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to;
                if(dis[v] > dis[u] + edge[i].val)
                {
                    dis[v] = dis[u] + edge[i].val;
                    Que.push(node(v, dis[v]));
                }
            }
        }
    }
};
using namespace Dijkstra;
inline void init()
{
    cnt = tot = 0;
    build(rt1, 0, 1, N, 0);
    build(rt2, 0, 1, N, 1);
    for(int i=1; i<=N; i++) addEddge(point_query(rt1, 1, N, i), point_query(rt2, 1, N, i), 0);
}
int main()
{
    scanf("%d%d%d", &N, &Q, &S);
    init();
    for(int i=1, op, v, l, r, w; i<=Q; i++)
    {
        scanf("%d", &op);
        switch (op)
        {
            case 1:
            {
                scanf("%d%d%d", &v, &l, &w);
                addEddge(point_query(rt2, 1, N, v), point_query(rt1, 1, N, l), w);
                break;
            }
            case 2:
            {
                scanf("%d%d%d%d", &v, &l, &r, &w);
                Range_query(rt1, 1, N, l, r, point_query(rt2, 1, N, v), w, 1);
                break;
            }
            default:
            {
                scanf("%d%d%d%d", &v, &l, &r, &w);
                Range_query(rt2, 1, N, l, r, point_query(rt1, 1, N, v), w, 0);
                break;
            }
        }
    }
    Dij();
    ll ans;
    for(int i=1; i<=N; i++)
    {
        ans = dis[point_query(rt2, 1, N, i)];
        printf("%lld%c", ans == BIG_INF ? -1 : ans, i == N ? '\n' : ' ');
    }
    return 0;
}