CF786B Legacy

Posted 2021-01-14 czxingchen

第一次写线段树优化建边。

根据本题的要求，我们可以建两棵线段树，然后在建$n$个点，然后把第一棵线段树上每一个点$(p, l, r)$（结点编号为$p$, 表示区间是$[l, r]$），连边$forall i in [l, r] (i, p, 0)$，与之相对，把第二棵线段树上的点$(p, l, r)$连边$forall i in [l, r]  (p, i, 0)$。

根据本题的特殊的连边条件限制，我们把一个区间拆成线段树上的$log$个点，然后在对应的点上连一下即可。

跑一遍最短路即可。

注意点数是$n + 2nlogn$，边数上限是$2nlogn + mlogn$，数组要开够。

时间复杂度$O(nlogn)$。

Code：

#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair <ll, int> pin;

const int N = 1e5 + 5;
const int M = 1.2e7 + 5;
const ll inf = 0x3f3f3f3f3f3f3f3f;

int n, m, tot = 0, head[N << 4];
ll dis[N << 4];
bool vis[N << 4];
vector <int> vec;

struct Edge {
    int to, nxt;
    ll val;
} e[M];

inline void add(int from, int to, ll val) {
    e[++tot].to = to;
    e[tot].val = val;
    e[tot].nxt = head[from];
    head[from] = tot;
}

template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for(; ch > ‘9‘|| ch < ‘0‘; ch = getchar())
        if(ch == ‘-‘) op = -1;
    for(; ch >= ‘0‘ && ch <= ‘9‘; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

namespace SegT {
    struct Node {
        int lc, rc;
    } s[N << 4];

    int root[2], nodeCnt;

    #define lc(p) s[p].lc
    #define rc(p) s[p].rc
    #define mid ((l + r) >> 1)

    void build(int &p, int l, int r, int type) {
        p = ++nodeCnt;

        for(int i = l; i <= r; i++)
            if(!type) add(i, p, 0LL);
            else add(p, i, 0LL);

        if(l == r) return;

        build(lc(p), l, mid, type);
        build(rc(p), mid + 1, r, type);
    }

    void getRan(int p, int l, int r, int x, int y) {
        if(x <= l && y >= r) {
            vec.push_back(p);
            return;
        }

        if(x <= mid) getRan(lc(p), l, mid, x, y);
        if(y > mid) getRan(rc(p), mid + 1, r, x, y);
    }

} using namespace SegT;

priority_queue <pin> Q;
inline void dij(int st) {
    memset(dis, 0x3f, sizeof(dis));
    Q.push(pin(dis[st] = 0LL, st));
    for(; !Q.empty(); ) {
        int x = Q.top().second; Q.pop();
        if(vis[x]) continue;
        vis[x] = 1;
        for(int i = head[x]; i; i = e[i].nxt) {
            int y = e[i].to;
            if(dis[y] > dis[x] + e[i].val) {
                dis[y] = dis[x] + e[i].val;
                Q.push(pin(-dis[y], y));
            }
        }
    }
}

int main() {
    read(n), read(m);
    int st; read(st);

    nodeCnt = n;
    build(root[0], 1, n, 0), build(root[1], 1, n, 1);
    for(int op, i = 1; i <= m; i++) {
        read(op);

        if(op == 1) {
            int x, y; ll v;
            read(x), read(y), read(v);
            add(x, y, v);
        }

        if(op == 2) {
            int x, l, r; ll v;
            read(x), read(l), read(r), read(v);

            vec.clear();
            getRan(root[1], 1, n, l, r);
            int vecSiz = vec.size();
            for(int j = 0; j < vecSiz; j++) {
                int y = vec[j];
                add(x, y, v);
            }
        }

        if(op == 3) {
            int y, l, r; ll v;
            read(y), read(l), read(r), read(v);

            vec.clear();
            getRan(root[0], 1, n, l, r);
            int vecSiz = vec.size();
            for(int j = 0; j < vecSiz; j++) {
                int x = vec[j];
                add(x, y, v);
            }
        }
    }

    dij(st);

    for(int i = 1; i <= n; i++)
        printf("%lld ", dis[i] == inf ? -1 : dis[i]);
    printf("
");

    return 0;
}

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