传送门

Sol

先正反两遍(Dijsktra)算出经过某个点的(S)到(T)的最短路条数(F)

满足条件一就是要满足(F(A)+F(B)=F(T))

条件二

标算比较简单

直接(bitset)存储不能到达它的和它不能到的点

然后开(map)把所有相同的(F(B))变成(bitset)

然后每次枚举(A)，就直接查表然后用(bitset)里的(count)就好了

注意如果(ST)不连通输出(C_n^2)

空间(1GB)就可以过

注意！！！

(AC)的同志不要以为真的(AC)了！！！
数据真的水
(S)到(T)的路径条数不会大于(1)
然后我自己乱打一个，样例都没过就(AC)了

标算代码

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
const int _(50005);
typedef long long ll;

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, cnt, S, T, first[_], d[_], vis[_];
struct Edge{
    int to, next, w;
} edge[_ << 1];
ll dis[2][_], f[2][_], F[_], ans;
struct Point{
    int u; ll d;

    IL int operator <(RG Point B) const{
        return d > B.d;
    }
};
priority_queue <Point> Q;
queue <int> Que;
bitset <50000> G[2][_];
map <ll, bitset <50000> > M;

IL void Add_Graph(RG int u, RG int v, RG int w){
    edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
}

IL void Dijkstra(RG int op){
    for(RG int i = 1; i <= n; ++i) dis[op][i] = (ll)1e18, vis[i] = 0;
    if(!op) f[op][S] = 1, dis[op][S] = 0, Q.push((Point){S, 0});
    else f[op][T] = 1, dis[op][T] = 0, Q.push((Point){T, 0});
    while(!Q.empty()){
        RG Point x = Q.top(); Q.pop();
        if(vis[x.u]) continue;
        vis[x.u] = 1;
        for(RG int e = first[x.u]; e != -1; e = edge[e].next){
            RG int v = edge[e].to, w = edge[e].w;
            if(x.d + w < dis[op][v]){
                dis[op][v] = x.d + w, f[op][v] = f[op][x.u];
                Q.push((Point){v, x.d + w});
            }
            else if(x.d + w == dis[op][v]) f[op][v] += f[op][x.u];
        }
    }
}

IL int Check(RG int x, RG ll w, RG int y, RG int op){
    return dis[op][x] + w + dis[!op][y] == dis[0][T];
}

IL void TopSort(RG int op){
    for(RG int i = 1; i <= n; ++i)
        for(RG int e = first[i]; e != -1; e = edge[e].next){
            RG int v = edge[e].to, w = edge[e].w;
            if(Check(i, w, v, op)) ++d[v];
        }
    for(RG int i = 1; i <= n; ++i) if(!d[i]) Que.push(i);
    while(!Que.empty()){
        RG int u = Que.front(); Que.pop();
        for(RG int e = first[u]; e != -1; e = edge[e].next){
            RG int v = edge[e].to, w = edge[e].w;
            if(!Check(u, w, v, op)) continue;
            G[op][v] &= G[op][u];
            if(!--d[v]) Que.push(v);
        }
    }
}

int main(RG int argc, RG char* argv[]){
    Fill(first, -1), n = Input(), m = Input(), S = Input(), T = Input();
    for(RG int i = 1; i <= m; ++i){
        RG int u = Input(), v = Input(), w = Input();
        Add_Graph(u, v, w), Add_Graph(v, u, w);
    }
    Dijkstra(0);
    if(!f[0][T]) return printf("%lld
", 1LL * n * (n - 1) >> 1), 0;
    Dijkstra(1);
    for(RG int i = 1; i <= n; ++i){
        G[0][i].set(), G[1][i].set();
        G[0][i][0] = G[0][i][i] = G[1][i][0] = G[1][i][i] = 0;
    }
    TopSort(0), TopSort(1);
    for(RG int i = 1; i <= n; ++i)
        if(dis[0][i] + dis[1][i] == dis[0][T]) F[i] = f[0][i] * f[1][i];
    for(RG int i = 1; i <= n; ++i) M[F[i]].set(i);
    for(RG int i = 1; i <= n; ++i)
        ans += (M[F[T] - F[i]] & G[0][i] & G[1][i]).count();
    return printf("%lld
", ans >> 1), 0;
}

然而Bzoj的空间变成了512MB

TAT

然后还可以这样做

首先可以枚举一条最短路上的点(i)

然后枚举不在这条最短路上的点再来统计答案

把最短路弄出来到一个数组中

然后不在这条路上的点的合法的(i)是这个数组的下标的区间([l, r])

然后可以正反两边拓扑排序求出这个区间

为什么是一个区间

首先前提是这个点走到(j)，然后(j)到(T)是最短路

• 如果(j)能走到这个点，那么(j)的所有前驱都能到这个点

• 如果这个点能走到(j)，那么这个点也能走到(j)的所有后继

然后枚举最短路上的点，每次到(i)时，就把左端点是(i)的加进(map)

然后把右端点是(i)的从(map)中减掉

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
const int _(50005);
typedef long long ll;

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, cnt, S, T, first[_], d[_], vis[_], l[_], r[_], p[_], len, pre[_];
struct Edge{
    int to, next, w;
} edge[_ << 1];
ll dis[2][_], f[2][_], F[_], ans;
struct Point{
    int u; ll d;

    IL int operator <(RG Point B) const{
        return d > B.d;
    }
};
queue <int> Que;
map <ll, int> M;
priority_queue <Point> Q;
vector <int> al[_], ar[_];

IL void Add_Graph(RG int u, RG int v, RG int w){
    edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
}

IL void Dijkstra(RG int op){
    for(RG int i = 1; i <= n; ++i) dis[op][i] = (ll)1e18, vis[i] = 0;
    if(!op) f[op][S] = 1, dis[op][S] = 0, Q.push((Point){S, 0});
    else f[op][T] = 1, dis[op][T] = 0, Q.push((Point){T, 0});
    while(!Q.empty()){
        RG Point x = Q.top(); Q.pop();
        if(vis[x.u]) continue;
        vis[x.u] = 1;
        for(RG int e = first[x.u]; e != -1; e = edge[e].next){
            RG int v = edge[e].to, w = edge[e].w;
            if(x.d + w < dis[op][v]){
                dis[op][v] = x.d + w, f[op][v] = f[op][x.u];
                Q.push((Point){v, x.d + w});
				if(op) pre[v] = x.u;
            }
            else if(x.d + w == dis[op][v]) f[op][v] += f[op][x.u];
        }
    }
}

IL int Check(RG int x, RG ll w, RG int y, RG int op){
    return dis[op][x] + w + dis[!op][y] == dis[0][T];
}

IL void TopSort(RG int op){
    for(RG int i = 1; i <= n; ++i)
        for(RG int e = first[i]; e != -1; e = edge[e].next){
            RG int v = edge[e].to, w = edge[e].w;
            if(Check(i, w, v, op)) ++d[v];
        }
    for(RG int i = 1; i <= n; ++i) if(!d[i]) Que.push(i);
    while(!Que.empty()){
        RG int u = Que.front(); Que.pop();
        for(RG int e = first[u]; e != -1; e = edge[e].next){
            RG int v = edge[e].to, w = edge[e].w;
            if(!Check(u, w, v, op)) continue;
			op ? r[v] = min(r[v], r[u]) : l[v] = max(l[v], l[u]);
            if(!--d[v]) Que.push(v);
        }
    }
}

int main(RG int argc, RG char* argv[]){
    Fill(first, -1), n = Input(), m = Input(), S = Input(), T = Input();
    for(RG int i = 1; i <= m; ++i){
        RG int u = Input(), v = Input(), w = Input();
        Add_Graph(u, v, w), Add_Graph(v, u, w);
    }
    Dijkstra(0);
    if(!f[0][T]) return printf("%lld
", 1LL * n * (n - 1) >> 1), 0;
    Dijkstra(1);
	for(RG int i = S; i; i = pre[i]) p[++len] = i, l[i] = len + 1, r[i] = len - 1;
	for(RG int i = 1; i <= n; ++i) if(!(l[i] + r[i])) l[i] = 1, r[i] = len;
	TopSort(0), TopSort(1);
    for(RG int i = 1; i <= n; ++i){
        if(dis[0][i] + dis[1][i] == dis[0][T]) F[i] = f[0][i] * f[1][i];
		if(l[i] > r[i]) continue;
		al[l[i]].push_back(i), ar[r[i]].push_back(i);
	}
	for(RG int i = 1; i <= len; ++i){
		for(RG int l = al[i].size(), j = 0; j < l; ++j) ++M[F[al[i][j]]];
		ans += M[F[T] - F[p[i]]];
		for(RG int l = ar[i].size(), j = 0; j < l; ++j) --M[F[ar[i][j]]];
	}
    return printf("%lld
", ans), 0;
}