BZOJ 3955 Surely You Congest 解题报告

首先，我们可以求出源为 $1$ 号点的最短路图以及各个点到 $1$ 号点的最短路。

然后我们考虑那些距离不同的点，是一定不会发生拥堵现象的。

然后我们就只需要考虑那些距离相同的点，就相当于做一个最大流嘛。

假设考虑与 $1$ 号节点距离为 $d$ 的点，那怎么连边，怎么设置源和汇呢？

• 源为 $1$ 号节点，新开一个 $n+1$ 号节点作为汇。
• 对于所有满足 $dist(1, x) + w(x,y) = dist(1, y)$ 的 $x,y$ 建一条 $x\rightarrow y$ 的边，容量为 $1$。
• 如果某个点 $x$ 与 $1$ 号节点距离恰好为 $d$，建一条 $x\rightarrow T$ 的边，容量为这个点上车辆的数目。

然后把所有距离下的最大流加起来，就是答案了。

复杂度看起来有点高，不过加点优化应该还是能跑过去的。

我加了一个优化：如果与 $1$ 号节点距离为 $d$ 的车辆只有 $1$ 辆，那么最大流就是 $1$，就不用去跑网络流了。

感觉效果还不错。

 #include <cmath>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 #define C 2000 + 5
 #define N 50000 + 5
 #define M 400000 + 5
 #define SIZE 10000000 + 5
 #define INF 0x7fffffff

 int n, m, c, S, T, tot, _tot, ans;
 int A[C];
 int Head[N], _Head[N], Inq[N];
 int Dis[N], _Dis[N];
 int E[M][];
 int q[SIZE];

 struct Edge
 {
     int next, node, flow, w;
 }h[M], _h[M];

 inline void addedge(int u, int v, int fl, int w)
 {
     h[++ tot].next = Head[u], Head[u] = tot;
     h[tot].node = v, h[tot].flow = fl, h[tot].w = w;
     h[++ tot].next = Head[v], Head[v] = tot;
     h[tot].node = u, h[tot].flow = , h[tot].w = w;
 }

 inline bool SPFA(int S)
 {
     for (int i = S; i <= T; i ++)
         Dis[i] = INF, Inq[i] = ;
     int l = , r = ;
     Dis[S] = , q[] = S, Inq[S] = ;
     while (l <= r)
     {
         int z = q[l ++];
         Inq[z] = ;
         for (int i = Head[z]; i; i = h[i].next)
         {
             int d = h[i].node, p = h[i].flow, w = h[i].w;
             if (!p) continue ;
             if (Dis[d] > Dis[z] + w)
             {
                 Dis[d] = Dis[z] + w;
                 if (!Inq[d])
                 {
                     q[++ r] = d;
                     Inq[d] = r;
                 }
             }
             if (Inq[d] && Dis[d] < Dis[q[l]])
             {
                 int u = Inq[d], v = q[l];
                 q[l] = d, q[u] = v;
                 Inq[d] = l, Inq[v] = u;
             }
         }
     }
     return Dis[T] != INF;
 }

 inline void Copy()
 {
     _tot = tot;
     for (int i = S; i <= T; i ++)
         _Head[i] = Head[i], _Dis[i] = Dis[i];
     for (int i = ; i <= tot; i ++)
         _h[i] = h[i];
 }

 inline void Restore()
 {
     tot = _tot;
     for (int i = S; i <= T; i ++)
         Head[i] = _Head[i];
     for (int i = ; i <= _tot; i ++)
         h[i] = _h[i];
 }

 inline bool cmp(int u, int v)
 {
     return Dis[u] < Dis[v];
 }

 inline int dinic(int z, int inflow)
 {
     if (z == T || !inflow) return inflow;
     int ret = inflow, flow;
     for (int i = Head[z]; i; i = h[i].next)
     {
         int d = h[i].node, p = h[i].flow;
         if (Dis[d] != Dis[z] + ) continue ;
         flow = dinic(d, min(ret, p));
         ret -= flow;
         h[i].flow -= flow, h[i ^ ].flow += flow;
         if (!ret) return inflow;
     }
     if (ret == inflow) Dis[z] = -;
     return inflow - ret;
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
         freopen("3955.in", "r", stdin);
         freopen("3955.out", "w", stdout);
     #endif

     scanf("%d%d%d", &n, &m, &c);
     S = , T = n + ;
     for (int i = ; i <= m; i ++)
     {
         int u, v, w;
         scanf("%d%d%d", &u ,&v, &w);
         E[i][] = u, E[i][] = v, E[i][] = w;
         addedge(u, v, , w);
         addedge(v, u, , w);
     }
     SPFA();
     for (int i = ; i <= c; i ++)
         scanf("%d", A + i);
     sort(A + , A + c + , cmp);
     tot = ;
     memset(Head, , sizeof(Head));
     for (int i = ; i <= m; i ++)
     {
         if (Dis[E[i][]] + E[i][] == Dis[E[i][]])
             addedge(E[i][], E[i][], , );
         if (Dis[E[i][]] + E[i][] == Dis[E[i][]])
             addedge(E[i][], E[i][], , );
     }
     Copy();
     int l = , r;
     for (; l <= c; l = r + )
     {
         for (r = l; r < c && _Dis[A[r + ]] == _Dis[A[l]]; r ++) ;
         if (r == l) ans ++;
         else
         {
             Restore();
             for (int i = l; i <= r; i ++)
                 addedge(A[i], T, , );
             while (SPFA(S))
                 ans += dinic(S, INF);
         }
     }
     printf("%d\n", ans);

     #ifndef ONLINE_JUDGE
         fclose(stdin);
         fclose(stdout);
     #endif
     return ;
 }

3955_Gromah