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。

 #include <cstdio>

 #include <cstring>

 #define int long long

 inline int nextChar(void) {

     const int siz = ;

     static char buf[siz];

     static char *hd = buf + siz;

     static char *tl = buf + siz;

     if (hd == tl)

         fread(hd = buf, , siz, stdin);

     return *hd++;

 }

 inline int nextInt(void) {

     register int ret = ;

     register int neg = false;

     register int bit = nextChar();

     for (; bit < ; bit = nextChar())

         if (bit == '-')neg ^= true;

     for (; bit > ; bit = nextChar())

         ret = ret *  + bit - ;

     return neg ? -ret : ret;

 }

 const int inf = 2e18;

 const int siz = ;

 int n, m;

 struct edge {

     int x, y, w;

 }e[siz];

 int lim[siz];

 namespace shortestPath

 {

     int dis[][siz];

     int edges;

     int hd[siz];

     int to[siz];

     int nt[siz];

     int vl[siz];

     inline void add(int u, int v, int w) {

         nt[edges] = hd[u]; to[edges] = v; vl[edges] = w; hd[u] = edges++;

         nt[edges] = hd[v]; to[edges] = u; vl[edges] = w; hd[v] = edges++;

     }

     inline void spfa(int *d, int s) {

         static int que[siz];

         static int inq[siz];

         static int head, tail;

         memset(inq, , sizeof(inq));

         for (int i = ; i < siz; ++i)d[i] = inf;

         inq[que[head = d[s] = ] = s] = tail = ;

         while (head != tail) {

             int u = que[head++], v; inq[u] = ;

             for (int i = hd[u]; ~i; i = nt[i])

                 if (d[v = to[i]] > d[u] + vl[i]) {

                     d[v] = d[u] + vl[i];

                     if (!inq[v])inq[que[tail++] = v] = ;

                 }

         }

     }

     inline void solve(void) {

         memset(hd, -, sizeof(hd));

         for (int i = ; i <= m; ++i)

             add(e[i].x, e[i].y, e[i].w);

         spfa(dis[], );

         spfa(dis[], n);

     }

 }

 namespace networkFlow

 {

     int s, t;

     int edges;

     int hd[siz];

     int to[siz];

     int nt[siz];

     int fl[siz];

     inline void add(int u, int v, int f) {

         nt[edges] = hd[u]; to[edges] = v; fl[edges] = f; hd[u] = edges++;

         nt[edges] = hd[v]; to[edges] = u; fl[edges] = ; hd[v] = edges++;

     }

     int dep[siz];

     inline bool bfs(void) {

         static int que[siz], head, tail;

         memset(dep, , sizeof(dep));

         dep[que[head = ] = s] = tail = ;

         while (head != tail) {

             int u = que[head++], v;

             for (int i = hd[u]; ~i; i = nt[i])

                 if (fl[i] && !dep[v = to[i]])

                     dep[que[tail++] = v] = dep[u] + ;

         }

         return dep[t];

     }

     int lst[siz];

     int dfs(int u, int f) {

         if (u == t || !f)return f;

         int used = , flow, v;

         for (int i = lst[u]; ~i; i = nt[i])

             if (dep[v = to[i]] == dep[u] + ) {

                 flow = dfs(v, f - used < fl[i] ? f - used : fl[i]);

                 used += flow;

                 fl[i] -= flow;

                 fl[i^] += flow;

                 if (fl[i])lst[u] = i;

                 if (used == f)return f;

             }

         if (!used)dep[u] = ;

         return used;

     }

     inline int maxFlow(void) {

         int maxFlow = , newFlow;

         while (bfs()) {

             for (int i = s; i <= t; ++i)

                 lst[i] = hd[i];

             while (newFlow = dfs(s, inf))

                 maxFlow += newFlow;

         }

         return maxFlow;

     }

     inline void solve(void) {

         s = , t = (n + ) << ;

         memset(hd, -, sizeof(hd));

         add(s,  << , inf);

         add(n <<  | , t, inf);

         for (int i = ; i <= n; ++i)

             add(i << , i <<  | , lim[i]);

         for (int i = ; i <= m; ++i) {

             int x, y, d = shortestPath::dis[][n];

             x = shortestPath::dis[][e[i].x];

             y = shortestPath::dis[][e[i].y];

             if (x + y + e[i].w == d)

                 add(e[i].x <<  | , e[i].y << , inf);

             x = shortestPath::dis[][e[i].y];

             y = shortestPath::dis[][e[i].x];

             if (x + y + e[i].w == d)

                 add(e[i].y <<  | , e[i].x << , inf);

         }

         printf("%lld\n", maxFlow());

     }

 }

 signed main(void) {

     n = nextInt();

     m = nextInt();

     for (int i = ; i <= m; ++i)

         e[i].x = nextInt(),

         e[i].y = nextInt(),

         e[i].w = nextInt();

     for (int i = ; i <= n; ++i)

         lim[i] = nextInt();

     lim[] = lim[n] = inf;

     shortestPath::solve();

     networkFlow::solve();

 }

@Author: YouSiki