Description

题库链接

给出一张 \(n\) 个点 \(m\) 条边有向图，询问最多有多少条不同的路径从 \(1\) 到 \(n\) 并且路径长度和 \(\leq E\) 。

\(2\leq n\leq 5000,1\leq m\leq 200000,1\leq E\leq 10^7\)

Solution

由于要求最多多少条，我们有贪心的思想，选取尽可能短的路径不会差。

那么题目就变成了求这张图 \(1\rightarrow n\) 的最短的 \(ans\) 条路径，并且路径的长度和 \(\leq E\) 。

就变成了裸的 \(k\) 短路问题。

由 \(f(x)=g(x)+h(x)\) ，由于让路径尽可能短，那么估价函数 \(h(x)\) 就取点 \(x\) 到 \(n\) 的最短路就好了。

b站上这道题代码写的丑的 \(stl\) 会 \(MLE\) ，像我这种菜鸡肯定过不了，只能手写可并堆了...

Code

//It is made by Awson on 2018.3.1

#include <bits/stdc++.h>

#define LL long long

#define dob complex<double>

#define Abs(a) ((a) < 0 ? (-(a)) : (a))

#define Max(a, b) ((a) > (b) ? (a) : (b))

#define Min(a, b) ((a) < (b) ? (a) : (b))

#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

#define writeln(x) (write(x), putchar('\n'))

#define lowbit(x) ((x)&(-(x)))

using namespace std;

const int N = 5000, M = 200000, INF = ~0u>>1;

void read(int &x) {

    char ch; bool flag = 0;

    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());

    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());

    x *= 1-2*flag;

}

void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }

void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, m, vis[N+5]; double dist[N+5], e;

struct node {

    double f, g; int u;

    bool operator < (const node &b) const {return f > b.f; }

};

struct mergeable_tree {

    int ch[1300005][2], dist[1300005], pos, root, cnt; node k[1300005];

    queue<int>mem;

    int newnode(node x) {

    int o; if (!mem.empty()) o = mem.front(), mem.pop(); else o = ++pos;

    ch[o][0] = ch[o][1] = dist[o] = 0; k[o] = x; return o;

    }

    int merge(int a, int b) {

    if (!a || !b) return a+b;

    if (k[a] < k[b]) Swap(a, b);

    ch[a][1] = merge(ch[a][1], b);

    if (dist[ch[a][0]] < dist[ch[a][1]]) Swap(ch[a][0], ch[a][1]);

    dist[a] = dist[ch[a][1]]+1; return a;

    }

    node top() {return k[root]; }

    void push(node x) {root = merge(root, newnode(x)); ++cnt; }

    void pop() {mem.push(root); root = merge(ch[root][0], ch[root][1]); --cnt; }

    bool empty() {return !cnt; }

}Q;

struct Graph {

    struct tt {int to, next; double cost; }edge[M+5];

    int path[N+5], top;

    void add(int u, int v, double c) {edge[++top].to = v, edge[top].next = path[u], edge[top].cost = c, path[u] = top; }

    void SPFA() {

    for (int i = 1; i < n; i++) dist[i] = INF;

    queue<int>Q; vis[n] = 1; Q.push(n);

    while (!Q.empty()) {

        int u = Q.front(); Q.pop(); vis[u] = 0;

        for (int i = path[u]; i; i = edge[i].next)

        if (dist[edge[i].to] > dist[u]+edge[i].cost) {

            dist[edge[i].to] = dist[u]+edge[i].cost;

            if (!vis[edge[i].to]) Q.push(edge[i].to), vis[edge[i].to] = 1;

        }

    }

    }

    int Astar() {

    int ans = 0;

    node t, tt; t.f = dist[1], t.g = 0, t.u = 1;

    Q.push(t);

    while (!Q.empty()) {

        t = Q.top(); Q.pop();

        if (t.u == n) {if (e >= t.f) {ans++, e -= t.f; continue; } else break; }

        for (int i = path[t.u]; i; i = edge[i].next) {

        tt.g = t.g+edge[i].cost, tt.u = edge[i].to, tt.f = tt.g+dist[edge[i].to];

        Q.push(tt);

        }

    }

    return ans;

    }

}g1, g2;

void work() {

    read(n), read(m); scanf("%lf", &e); int u, v; double c;

    for (int i = 1; i <= m; i++) {

    read(u), read(v), scanf("%lf", &c), g1.add(u, v, c), g2.add(v, u, c);

    }

    g2.SPFA(); writeln(g1.Astar());

}

int main() {

    work(); return 0;

}