[笔记]网络流
一些资料：最详细网络流建模基础
模板
最大流Dinic
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
const int inf = 0x3fffffff;
const int N = 10005, M = 100005;
int n, m, s, t, maxflow;
int d[N];
struct Edge {
int Nxt, v, flow;
}e[M << 1];
int h[N], p = 1;
void add(int u, int v, int z)
{
e[++p].Nxt = h[u]; e[p].v = v; e[p].flow = z; h[u] = p;
e[++p].Nxt = h[v]; e[p].v = u; e[p].flow = 0; h[v] = p;
}
std::queue<int> q;
bool bfs()
{
memset(d, 0, sizeof(d));
while(!q.empty()) q.pop();
q.push(s); d[s] = 1;
while(!q.empty())
{
int u = q.front(); q.pop();
for(int i = h[u]; i; i = e[i].Nxt)
if(e[i].flow && !d[e[i].v])
{
q.push(e[i].v);
d[e[i].v] = d[u] + 1;
if(e[i].v == t) return true;
}
}
return false;
}
int dinic(int u, int flow)
{
if(u == t) return flow;
int rest = flow, k;
for(int i = h[u]; i && rest; i = e[i].Nxt)
if(e[i].flow && d[e[i].v] == d[u] + 1)
{
k = dinic(e[i].v, std::min(rest, e[i].flow));
if(!k) d[e[i].v] = 0;
e[i].flow -= k;
e[i ^ 1].flow += k;
rest -= k;
}
return flow - rest;
}
int main()
{
scanf("%d%d%d%d", &n, &m, &s, &t);
for(int i = 1, u, v, z; i <= m; i++) {
scanf("%d%d%d", &u, &v, &z);
add(u, v, z);
}
int flow = 0;
while(bfs())
while(flow = dinic(s, inf)) maxflow += flow;
printf("%d\n", maxflow);
return 0;
}
费用流EK
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
const int _ = 50005;
const int inf = 0x3f3f3f3f;
int n, m, s, t;
int ans, maxflow;
int dist[_], incf[_], pre[_], vis[_];
struct Edge { int Nxt, v, flow, cost; } e[_ << 1];
int h[_], p = 1;
void add(int u, int v, int f, int c)
{
e[++p].Nxt = h[u]; e[p].v = v; e[p].flow = f; e[p].cost = c; h[u] = p;
e[++p].Nxt = h[v]; e[p].v = u; e[p].flow = 0; e[p].cost = -c; h[v] = p;
}
std::queue<int> q;
bool spfa()
{
memset(dist, 0x3f, sizeof(dist));
memset(vis, 0, sizeof(vis));
q.push(s); dist[s] = 0; vis[s] = 1;
incf[s] = inf;
while(!q.empty())
{
int u = q.front(); q.pop();
vis[u] = 0;
for(int i = h[u]; i; i = e[i].Nxt)
{
if(!e[i].flow) continue;
int v = e[i].v;
if(dist[v] > dist[u] + e[i].cost)
{
dist[v] = dist[u] + e[i].cost;
incf[v] = std::min(incf[u], e[i].flow);
pre[v] = i;
if(!vis[v]) vis[v] = 1, q.push(v);
}
}
}
if(dist[t] == inf) return false;
else return true;
}
void update()
{
int u = t;
while(u != s)
{
int i = pre[u];
e[i].flow -= incf[t];
e[i ^ 1].flow += incf[t];
u = e[i ^ 1].v;
}
maxflow += incf[t];
ans += dist[t] * incf[t];
}
int main()
{
scanf("%d%d%d%d", &n, &m, &s, &t);
for(int i = 1, u, v, z, c; i <= m; i++) {
scanf("%d%d%d%d", &u, &v, &z, &c);
add(u, v, z, c);
}
while(spfa()) update();
printf("%d %d\n", maxflow, ans);
return 0;
}
Problem
神奇链上建图
算法：线性规划 or 有源汇上下界最小费用可行流可以无视
一开始幻想这样可以搞：
但其实很难限制条件并且流量并不能准确一对多去覆盖点。
所以是建成这样的：
但是为什么对呢，其实非常玄学，我觉得还是手动模拟一下不会证，然后感性理解。
模拟时可以发现\(inf-Ai\)其实对人数限制取反，代表要跑完这条边所有流量，流到\(0\)之前的流量是没有实际作用的，是为了可以先跳过这天，直到流量为\(0\)后，代表这条边需要开始选择人进行工作，而这条边需要的人数可以通过别的边反映出来，可以从红边即人补全这条链上为\(0\)的边无法通过的流量，通过最小费用最大流即可求出最小费用。
\(n+1->t\)这条边可以反映出当前还剩的未规划的最大人数，所以\(n+1->t\)流量为\(0\)时，代表之前最大流量流尽，即\((inf-A[i])_{max}\)，即\(A[i]_{min}\)。
神奇动态加边
[NOI2012]美食节
从算法到算术，使用小学数学并利用大量算术技巧计算点的遍号
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define R register
using namespace std;
const int N = 45, M = 105, sP = 805, _ = N + M * sP + 5;
const int inf = 0x3f3f3f3f;
int n, m, s, t, sum, tim[N][M], P[N];
struct Edge { int Next, v, flow, cost; } e[(N + N * M * sP + M * sP + 5) << 1];
int h[N + M * sP + 5], p = 1;
inline void add(R int u, R int v, int f, int c) {
e[++p].Next = h[u]; e[p].v = v; e[p].flow = f; e[p].cost = c; h[u] = p;
e[++p].Next = h[v]; e[p].v = u; e[p].flow = 0; e[p].cost = -c; h[v] = p;
}
int dist[_], incf[_], maxflow, ans, vis[_], pre[_];
queue<int> q;
bool spfa()
{
memset(vis, 0, sizeof(vis));
memset(dist, 0x3f, sizeof(dist));
dist[s] = 0; vis[s] = 1; q.push(s); incf[s] = inf;
while(!q.empty()) {
int u = q.front(); q.pop();
vis[u] = 0;
for(int i = h[u]; i; i = e[i].Next)
if(e[i].flow)
{
int v = e[i].v;
if(dist[v] > dist[u] + e[i].cost) {
dist[v] = dist[u] + e[i].cost;
incf[v] = min(incf[u], e[i].flow);
pre[v] = i;
if(!vis[v]) q.push(v), vis[v] = 1;
}
}
}
if(dist[t] == inf) return false;
else return true;
}
inline void update() {
R int u = t;
while(u != s) {
int i = pre[u];
e[i].flow -= incf[t];
e[i ^ 1].flow += incf[t];
u = e[i ^ 1].v;
}
maxflow += incf[t];
ans += dist[t] * incf[t];
}
int main()
{
scanf("%d%d", &n, &m);
for(R int i = 1; i <= n; i++) scanf("%d", &P[i]), sum += P[i];
s = 0, t = m * sum + n + 1;
for(R int i = 1; i <= n; i++)
for(R int j = 1; j <= m; j++)
scanf("%d", &tim[i][j]);
for(R int i = 1; i <= n; i++) add(s, i, P[i], 0);
for(R int i = n + 1; i <= n + m * sum; i += sum) add(i, t, 1, 0);
for(R int i = 1; i <= n; i++)
for(R int j = n + 1, t = 1; j <= n + m * sum; j += sum)
add(i, j, 1, tim[i][t++]);
while(spfa()) {
update();
int u = e[pre[t] ^ 1].v;
if((u - n) % sum == 0) break;
int v = u + 1;
int k = (v - n) % sum == 0 ? sum : (v - n) % sum;
int j = (v - n - k) / sum + 1;
add(v, t, 1, 0);
for(R int i = 1; i <= n; i++)
add(i, v, 1, k * tim[i][j]);
}
printf("%d\n", ans);
return 0;
}
最大权闭合子图
定理 最大权闭合子图权值=正权点之和-最小割
