Bzoj5109: [CodePlus 2017]大吉大利，晚上吃雞！
阿新 • • 發佈：2018-04-02
check class strong vector pre != truct count() href
題面
傳送門
Sol
先正反兩遍\(Dijsktra\)算出經過某個點的\(S\)到\(T\)的最短路條數\(F\)
滿足條件一就是要滿足\(F(A)+F(B)=F(T)\)
條件二
標算比較簡單
直接\(bitset\)存儲不能到達它的和它不能到的點
然後開\(map\)把所有相同的\(F(B)\)變成\(bitset\)
然後每次枚舉\(A\)，就直接查表然後用\(bitset\)裏的\(count\)就好了
註意如果\(ST\)不連通輸出\(C_n^2\)
空間\(1GB\)就可以過
註意！！！
\(AC\)的同誌不要以為真的\(AC\)了！！！
數據真的水
\(S\)到\(T\)的路徑條數不會大於\(1\)
然後我自己亂打一個，樣例都沒過就\(AC\)了
標算代碼
# 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' 然而Bzoj的空間變成了512MB
TAT
然後還可以這樣做
首先可以枚舉一條最短路上的點\(i\)
然後枚舉不在這條最短路上的點再來統計答案
把最短路弄出來到一個數組中
然後不在這條路上的點的合法的\(i\)是這個數組的下標的區間\([l, r]\)
然後可以正反兩邊拓撲排序求出這個區間
為什麽是一個區間
首先前提是這個點走到\(j\)，然後\(j\)到\(T\)是最短路
如果\(j\)能走到這個點，那麽\(j\)的所有前驅都能到這個點
如果這個點能走到\(j\)，那麽這個點也能走到\(j\)的所有後繼
然後枚舉最短路上的點，每次到\(i\)時，就把左端點是\(i\)的加進\(map\)
然後把右端點是\(i\)的從\(map\)中減掉
# 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\n", 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\n", ans), 0;
}Bzoj5109: [CodePlus 2017]大吉大利，晚上吃雞！