BZOJ 3931: [CQOI2015]网络吞吐量

3931: [CQOI2015]网络吞吐量

Time Limit: 10 Sec Memory Limit: 512 MB
Submit: 1555 Solved: 637
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Description

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。

Input

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。

Output

输出一个整数，为题目所求吞吐量。

Sample Input

7 10
1 2 2
1 5 2
2 4 1
2 3 3
3 7 1
4 5 4
4 3 1
4 6 1
5 6 2
6 7 1
1
100
20
50
20
60
1

Sample Output

HINT

对于100%的数据，n≤500，m≤100000，d,c≤10^9

Source

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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
", 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 }

@Author: YouSiki