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

70

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 = 1024;
static char buf[siz];
static char *hd = buf + siz;
static char *tl = buf + siz;
if (hd == tl)
fread(hd = buf, 1, siz, stdin);
return *hd++;
}

inline int nextInt(void) {
register int ret = 0;
register int neg = false;
register int bit = nextChar();
for (; bit < 48; bit = nextChar())
if (bit == '-')neg ^= true;
for (; bit > 47; bit = nextChar())
ret = ret * 10 + bit - 48;
return neg ? -ret : ret;
}

const int inf = 2e18;
const int siz = 1000005;

int n, m;

struct edge {
int x, y, w;
}e[siz];

int lim[siz];

namespace shortestPath
{
int dis[2][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, 0, sizeof(inq));
for (int i = 0; i < siz; ++i)d[i] = inf;
inq[que[head = d[s] = 0] = s] = tail = 1;
while (head != tail) {
int u = que[head++], v; inq[u] = 0;
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] = 1;
}
}
}

inline void solve(void) {
memset(hd, -1, sizeof(hd));
for (int i = 1; i <= m; ++i)
add(e[i].x, e[i].y, e[i].w);
spfa(dis[0], 1);
spfa(dis[1], 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] = 0; hd[v] = edges++;
}

int dep[siz];

inline bool bfs(void) {
static int que[siz], head, tail;
memset(dep, 0, sizeof(dep));
dep[que[head = 0] = s] = tail = 1;
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] + 1;
}
return dep[t];
}

int lst[siz];

int dfs(int u, int f) {
if (u == t || !f)return f;
int used = 0, flow, v;
for (int i = lst[u]; ~i; i = nt[i])
if (dep[v = to[i]] == dep[u] + 1) {
flow = dfs(v, f - used < fl[i] ? f - used : fl[i]);
used += flow;
fl[i] -= flow;
fl[i^1] += flow;
if (fl[i])lst[u] = i;
if (used == f)return f;
}
if (!used)dep[u] = 0;
return used;
}

inline int maxFlow(void) {
int maxFlow = 0, 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 = 0, t = (n + 1) << 1;
memset(hd, -1, sizeof(hd));
add(s, 1 << 1, inf);
add(n << 1 | 1, t, inf);
for (int i = 1; i <= n; ++i)
add(i << 1, i << 1 | 1, lim[i]);
for (int i = 1; i <= m; ++i) {
int x, y, d = shortestPath::dis[0]
;
x = shortestPath::dis[0][e[i].x];
y = shortestPath::dis[1][e[i].y];
if (x + y + e[i].w == d)
add(e[i].x << 1 | 1, e[i].y << 1, inf);
x = shortestPath::dis[0][e[i].y];
y = shortestPath::dis[1][e[i].x];
if (x + y + e[i].w == d)
add(e[i].y << 1 | 1, e[i].x << 1, inf);
}
printf("%lld\n", maxFlow());
}
}

signed main(void) {
n = nextInt();
m = nextInt();
for (int i = 1; i <= m; ++i)
e[i].x = nextInt(),
e[i].y = nextInt(),
e[i].w = nextInt();
for (int i = 1; i <= n; ++i)
lim[i] = nextInt();
lim[1] = lim
= inf;
shortestPath::solve();
networkFlow::solve();
}

@Author: YouSiki