路由是指通过计算机网络把信息从源地址传输到目的地址的活动,也是计算机网络设计中的重点和难点。网络中实现路由转发的硬件设备称为路由器。为了使数据包最快的到达目的地,路由器需要选择最优的路径转发数据包。例如在常用的路由算法OSPF(开放式最短路径优先)中,路由器会使用经典的Dijkstra算法计算最短路径,然后尽量沿最短路径转发数据包。现在,若已知一个计算机网络中各路由器间的连接情况,以及各个路由器的最大吞吐量(即每秒能转发的数据包数量),假设所有数据包一定沿最短路径转发,试计算从路由器1到路由器n的网络的最大吞吐量。计算中忽略转发及传输的时间开销,不考虑链路的带宽限制,即认为数据包可以瞬间通过网络。路由器1到路由器n作为起点和终点,自身的吞吐量不用考虑,网络上也不存在将1和n直接相连的链路。
输入文件第一行包含两个空格分开的正整数n和m,分别表示路由器数量和链路的数量。网络中的路由器使用1到n编号。接下来m行,每行包含三个空格分开的正整数a、b和d,表示从路由器a到路由器b存在一条距离为d的双向链路。 接下来n行,每行包含一个正整数c,分别给出每一个路由器的吞吐量。
对于100%的数据,n≤500,m≤100000,d,c≤10^9
分别从1和n点做单源最短路,即可求出哪些边出现在了从1到n的最短路上(最短路不一定唯一)。将这些边加入网络流中,对于原本的点拆点,入点向出点连限制吞吐量容量的边,跑最大流。注意Int64。
1 #include <cstdio>
2 #include <cstring>
3
4 #define int long long
5
6 inline int nextChar(void) {
7 const int siz = 1024;
8 static char buf[siz];
9 static char *hd = buf + siz;
10 static char *tl = buf + siz;
11 if (hd == tl)
12 fread(hd = buf, 1, siz, stdin);
13 return *hd++;
14 }
15
16 inline int nextInt(void) {
17 register int ret = 0;
18 register int neg = false;
19 register int bit = nextChar();
20 for (; bit < 48; bit = nextChar())
21 if (bit == ‘-‘)neg ^= true;
22 for (; bit > 47; bit = nextChar())
23 ret = ret * 10 + bit - 48;
24 return neg ? -ret : ret;
25 }
26
27 const int inf = 2e18;
28 const int siz = 1000005;
29
30 int n, m;
31
32 struct edge {
33 int x, y, w;
34 }e[siz];
35
36 int lim[siz];
37
38 namespace shortestPath
39 {
40 int dis[2][siz];
41
42 int edges;
43 int hd[siz];
44 int to[siz];
45 int nt[siz];
46 int vl[siz];
47
48 inline void add(int u, int v, int w) {
49 nt[edges] = hd[u]; to[edges] = v; vl[edges] = w; hd[u] = edges++;
50 nt[edges] = hd[v]; to[edges] = u; vl[edges] = w; hd[v] = edges++;
51 }
52
53 inline void spfa(int *d, int s) {
54 static int que[siz];
55 static int inq[siz];
56 static int head, tail;
57 memset(inq, 0, sizeof(inq));
58 for (int i = 0; i < siz; ++i)d[i] = inf;
59 inq[que[head = d[s] = 0] = s] = tail = 1;
60 while (head != tail) {
61 int u = que[head++], v; inq[u] = 0;
62 for (int i = hd[u]; ~i; i = nt[i])
63 if (d[v = to[i]] > d[u] + vl[i]) {
64 d[v] = d[u] + vl[i];
65 if (!inq[v])inq[que[tail++] = v] = 1;
66 }
67 }
68 }
69
70 inline void solve(void) {
71 memset(hd, -1, sizeof(hd));
72 for (int i = 1; i <= m; ++i)
73 add(e[i].x, e[i].y, e[i].w);
74 spfa(dis[0], 1);
75 spfa(dis[1], n);
76 }
77 }
78
79 namespace networkFlow
80 {
81 int s, t;
82 int edges;
83 int hd[siz];
84 int to[siz];
85 int nt[siz];
86 int fl[siz];
87
88 inline void add(int u, int v, int f) {
89 nt[edges] = hd[u]; to[edges] = v; fl[edges] = f; hd[u] = edges++;
90 nt[edges] = hd[v]; to[edges] = u; fl[edges] = 0; hd[v] = edges++;
91 }
92
93 int dep[siz];
94
95 inline bool bfs(void) {
96 static int que[siz], head, tail;
97 memset(dep, 0, sizeof(dep));
98 dep[que[head = 0] = s] = tail = 1;
99 while (head != tail) {
100 int u = que[head++], v;
101 for (int i = hd[u]; ~i; i = nt[i])
102 if (fl[i] && !dep[v = to[i]])
103 dep[que[tail++] = v] = dep[u] + 1;
104 }
105 return dep[t];
106 }
107
108 int lst[siz];
109
110 int dfs(int u, int f) {
111 if (u == t || !f)return f;
112 int used = 0, flow, v;
113 for (int i = lst[u]; ~i; i = nt[i])
114 if (dep[v = to[i]] == dep[u] + 1) {
115 flow = dfs(v, f - used < fl[i] ? f - used : fl[i]);
116 used += flow;
117 fl[i] -= flow;
118 fl[i^1] += flow;
119 if (fl[i])lst[u] = i;
120 if (used == f)return f;
121 }
122 if (!used)dep[u] = 0;
123 return used;
124 }
125
126 inline int maxFlow(void) {
127 int maxFlow = 0, newFlow;
128 while (bfs()) {
129 for (int i = s; i <= t; ++i)
130 lst[i] = hd[i];
131 while (newFlow = dfs(s, inf))
132 maxFlow += newFlow;
133 }
134 return maxFlow;
135 }
136
137 inline void solve(void) {
138 s = 0, t = (n + 1) << 1;
139 memset(hd, -1, sizeof(hd));
140 add(s, 1 << 1, inf);
141 add(n << 1 | 1, t, inf);
142 for (int i = 1; i <= n; ++i)
143 add(i << 1, i << 1 | 1, lim[i]);
144 for (int i = 1; i <= m; ++i) {
145 int x, y, d = shortestPath::dis[0][n];
146 x = shortestPath::dis[0][e[i].x];
147 y = shortestPath::dis[1][e[i].y];
148 if (x + y + e[i].w == d)
149 add(e[i].x << 1 | 1, e[i].y << 1, inf);
150 x = shortestPath::dis[0][e[i].y];
151 y = shortestPath::dis[1][e[i].x];
152 if (x + y + e[i].w == d)
153 add(e[i].y << 1 | 1, e[i].x << 1, inf);
154 }
155 printf("%lld\n", maxFlow());
156 }
157 }
158
159 signed main(void) {
160 n = nextInt();
161 m = nextInt();
162 for (int i = 1; i <= m; ++i)
163 e[i].x = nextInt(),
164 e[i].y = nextInt(),
165 e[i].w = nextInt();
166 for (int i = 1; i <= n; ++i)
167 lim[i] = nextInt();
168 lim[1] = lim[n] = inf;
169 shortestPath::solve();
170 networkFlow::solve();
171 }