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。
1 #include <cmath>
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7 typedef long long LL;
8 #define C 2000 + 5
9 #define N 50000 + 5
10 #define M 400000 + 5
11 #define SIZE 10000000 + 5
12 #define INF 0x7fffffff
13
14 int n, m, c, S, T, tot, _tot, ans;
15 int A[C];
16 int Head[N], _Head[N], Inq[N];
17 int Dis[N], _Dis[N];
18 int E[M][3];
19 int q[SIZE];
20
21 struct Edge
22 {
23 int next, node, flow, w;
24 }h[M], _h[M];
25
26 inline void addedge(int u, int v, int fl, int w)
27 {
28 h[++ tot].next = Head[u], Head[u] = tot;
29 h[tot].node = v, h[tot].flow = fl, h[tot].w = w;
30 h[++ tot].next = Head[v], Head[v] = tot;
31 h[tot].node = u, h[tot].flow = 0, h[tot].w = w;
32 }
33
34 inline bool SPFA(int S)
35 {
36 for (int i = S; i <= T; i ++)
37 Dis[i] = INF, Inq[i] = 0;
38 int l = 1, r = 1;
39 Dis[S] = 0, q[1] = S, Inq[S] = 1;
40 while (l <= r)
41 {
42 int z = q[l ++];
43 Inq[z] = 0;
44 for (int i = Head[z]; i; i = h[i].next)
45 {
46 int d = h[i].node, p = h[i].flow, w = h[i].w;
47 if (!p) continue ;
48 if (Dis[d] > Dis[z] + w)
49 {
50 Dis[d] = Dis[z] + w;
51 if (!Inq[d])
52 {
53 q[++ r] = d;
54 Inq[d] = r;
55 }
56 }
57 if (Inq[d] && Dis[d] < Dis[q[l]])
58 {
59 int u = Inq[d], v = q[l];
60 q[l] = d, q[u] = v;
61 Inq[d] = l, Inq[v] = u;
62 }
63 }
64 }
65 return Dis[T] != INF;
66 }
67
68 inline void Copy()
69 {
70 _tot = tot;
71 for (int i = S; i <= T; i ++)
72 _Head[i] = Head[i], _Dis[i] = Dis[i];
73 for (int i = 2; i <= tot; i ++)
74 _h[i] = h[i];
75 }
76
77 inline void Restore()
78 {
79 tot = _tot;
80 for (int i = S; i <= T; i ++)
81 Head[i] = _Head[i];
82 for (int i = 2; i <= _tot; i ++)
83 h[i] = _h[i];
84 }
85
86 inline bool cmp(int u, int v)
87 {
88 return Dis[u] < Dis[v];
89 }
90
91 inline int dinic(int z, int inflow)
92 {
93 if (z == T || !inflow) return inflow;
94 int ret = inflow, flow;
95 for (int i = Head[z]; i; i = h[i].next)
96 {
97 int d = h[i].node, p = h[i].flow;
98 if (Dis[d] != Dis[z] + 1) continue ;
99 flow = dinic(d, min(ret, p));
100 ret -= flow;
101 h[i].flow -= flow, h[i ^ 1].flow += flow;
102 if (!ret) return inflow;
103 }
104 if (ret == inflow) Dis[z] = -1;
105 return inflow - ret;
106 }
107
108 int main()
109 {
110 #ifndef ONLINE_JUDGE
111 freopen("3955.in", "r", stdin);
112 freopen("3955.out", "w", stdout);
113 #endif
114
115 scanf("%d%d%d", &n, &m, &c);
116 S = 1, T = n + 1;
117 for (int i = 1; i <= m; i ++)
118 {
119 int u, v, w;
120 scanf("%d%d%d", &u ,&v, &w);
121 E[i][0] = u, E[i][1] = v, E[i][2] = w;
122 addedge(u, v, 1, w);
123 addedge(v, u, 1, w);
124 }
125 SPFA(1);
126 for (int i = 1; i <= c; i ++)
127 scanf("%d", A + i);
128 sort(A + 1, A + c + 1, cmp);
129 tot = 1;
130 memset(Head, 0, sizeof(Head));
131 for (int i = 1; i <= m; i ++)
132 {
133 if (Dis[E[i][0]] + E[i][2] == Dis[E[i][1]])
134 addedge(E[i][0], E[i][1], 1, 1);
135 if (Dis[E[i][1]] + E[i][2] == Dis[E[i][0]])
136 addedge(E[i][1], E[i][0], 1, 1);
137 }
138 Copy();
139 int l = 1, r;
140 for (; l <= c; l = r + 1)
141 {
142 for (r = l; r < c && _Dis[A[r + 1]] == _Dis[A[l]]; r ++) ;
143 if (r == l) ans ++;
144 else
145 {
146 Restore();
147 for (int i = l; i <= r; i ++)
148 addedge(A[i], T, 1, 1);
149 while (SPFA(S))
150 ans += dinic(S, INF);
151 }
152 }
153 printf("%d\n", ans);
154
155 #ifndef ONLINE_JUDGE
156 fclose(stdin);
157 fclose(stdout);
158 #endif
159 return 0;
160 }
3955_Gromah
时间： 2024-11-01 16:15:33