[AtCoder arc090E]Avoiding Collision

Description

题库链接

给出一张 $N$ 个节点， $M$ 条边的无向图，给出起点 $S$ 和终点 $T$ 。询问两个人分别从 $S$ 和 $T$ 出发，走最短路不相遇的方案数。

$1 \leq N \leq 100,000,1 \leq M \leq 200,000$

Solution

记最短路长度为 $D$ ，从 $S$ 出发走过的路程为 $d_1$ ，从 $T$ 出发的走过的路程为 $d_2$ 。值得注意的是走最短路相遇只会出现两种情况：

1. 在某个节点相遇，此时 $d_1=d_2=\frac{D}{2}$ ；
2. 在某条边上相遇，记这条边为 $e$ ，边权为 $c$ ，连接 $u,v$ 两个节点。此时 ${d_1}_u+{d_2}_v+c=D$ ，且 $2{d_1}_u < D, 2{d_2}_v < D$

那么我们跑一遍最短路计数，用总方案数减去上述不合法的情况即可。

Code

//It is made by Awson on 2018.2.2
#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 = 100000;
const int yzd = 1e9+7;
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, s, t, u, v, c;
struct tt {int to, next, cost; }edge[(N<<2)+5];
int path[N+5], top;
int ans1[N+5], ans2[N+5], vis[N+5], in[N+5];
LL dist1[N+5], dist2[N+5];
queue<int>Q;
vector<int>to[N+5];

void add(int u, int v, int c) {
    edge[++top].to = v, edge[top].cost = c, edge[top].next = path[u], path[u] = top;
}
void SPFA(int s, LL *dist) {
    dist[s] = 0; Q.push(s); vis[s] = 1;
    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]) {
            vis[edge[i].to] = 1; Q.push(edge[i].to);
        }
        }
    }
}
void topsort(int s, LL *dist, int *ans) {
    for (int u = 1; u <= n; u++) {
    to[u].clear();
    for (int i = path[u]; i; i = edge[i].next)
        if (dist[edge[i].to] == dist[u]+edge[i].cost) to[u].push_back(edge[i].to), ++in[edge[i].to];
    }
    ans[s] = 1; Q.push(s);
    while (!Q.empty()) {
    int u = Q.front(), size = to[u].size(); Q.pop();
    for (int i = 0; i < size; i++) {
        if (--in[to[u][i]] == 0) Q.push(to[u][i]); (ans[to[u][i]] += ans[u]) %= yzd;
    }
    }
}
void work() {
    read(n), read(m), read(s), read(t);
    for (int i = 1; i <= m; i++) {
    read(u), read(v), read(c); add(u, v, c); add(v, u, c);
    }
    memset(dist1, 127/3, sizeof(dist1));
    SPFA(s, dist1); topsort(s, dist1, ans1);
    memset(dist2, 127/3, sizeof(dist2));
    SPFA(t, dist2); topsort(t, dist2, ans2);
    int ans = 1ll*ans1[t]*ans1[t]%yzd;
    for (int u = 1; u <= n; u++) {
    if (dist1[u] == dist2[u] && dist2[u]*2 == dist1[t]) (ans -= 1ll*ans1[u]*ans1[u]%yzd*ans2[u]%yzd*ans2[u]%yzd) %= yzd;
    for (int i = path[u]; i; i = edge[i].next)
        if (dist1[u]+edge[i].cost+dist2[edge[i].to] == dist1[t] && dist1[u]*2 < dist1[t] && dist2[edge[i].to]*2 < dist1[t])
        (ans -= 1ll*ans1[u]*ans1[u]%yzd*ans2[edge[i].to]%yzd*ans2[edge[i].to]%yzd) %= yzd;
    }
    writeln((ans+yzd)%yzd);
}
int main() {
    work();
    return 0;
}