题目来源
最大流
最大流问题是网络流的经典类型之一，用处广泛，个人认为网络流问题最具特点的操作就是建反向边，这样相当于给了反悔的机会，不断地求增广路的，最终得到最大流
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<string>
#include<fstream>
#include<vector>
#include<stack>
#include <map>
#include <iomanip>
#define bug cout << "**********" << endl
#define show(x,y) cout<<"["<<x<<","<<y<<"] "
//#define LOCAL = 1;
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll mod = 1e6 + 3;
const int Max = 1e5 + 10;
struct Edge {
int to, next, flow; //flow记录这条边当前的边残量
}edge[Max << 1];
int n, m, s, t;
int head[Max], tot;
bool vis[Max];
void init()
{
memset(head, -1, sizeof(head));tot = 0;
}
void add(int u, int v, int flow)
{
edge[tot].to = v;
edge[tot].flow = flow;
edge[tot].next = head[u];
head[u] = tot++;
}
//向图中增加一条容量为exp的边（增广路）
int dfs(int u,int exp)
{
if (u == t) return exp; //到达汇点，当前水量全部注入
vis[u] = true; //表示已经到了过了
for(int i = head[u] ; i != -1 ;i = edge[i].next)
{
int v = edge[i].to;
if(!vis[v] && edge[i].flow > 0)
{
int flow = dfs(v, min(exp, edge[i].flow));
if(flow > 0) //形成了增广路
{
edge[i].flow -= flow;
edge[i ^ 1].flow += flow;
return flow;
}
}
}
return 0; //无法形成增广路的情况
}
//求最大流
int max_flow()
{
int flow = 0;
while(true)
{
memset(vis, 0, sizeof(vis));
int part_flow = dfs(s, inf);
if (part_flow == 0) return flow;
flow += part_flow;
}
}
int main()
{
#ifdef LOCAL
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
while (scanf("%d%d%d%d", &n, &m, &s, &t) != EOF)
{
init();
for (int i = 1, u, v, flow;i <= m; i++)
{
scanf("%d%d%d", &u, &v, &flow);
add(u, v, flow);add(v, u, 0);
}
printf("%d\n", max_flow());
}
return 0;
}
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<string>
#include<fstream>
#include<vector>
#include<stack>
#include <map>
#include <iomanip>
#define bug cout << "**********" << endl
#define show(x,y) "["<<x<<","<<y<<"]"
//#define LOCAL = 1;
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll mod = 1e6 + 3;
const int Max = 1e5 + 10;
struct Edge {
int to, next, flow; //flow记录这条边当前的边残量
}edge[Max << 1];
int n, m, s, t;
int head[Max], tot;
int dis[Max];
void init()
{
memset(head, -1, sizeof(head));tot = 0;
}
void add(int u, int v, int flow)
{
edge[tot].to = v;
edge[tot].flow = flow;
edge[tot].next = head[u];
head[u] = tot++;
}
bool bfs() //判断图是否连通
{
queue<int>q;
memset(dis, -1, sizeof(dis));
dis[s] = 0;
q.push(s);
while (!q.empty())
{
int u = q.front();q.pop();
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (dis[v] == -1 && edge[i].flow > 0) //可以借助边i到达新的结点
{
dis[v] = dis[u] + 1; //求顶点到源点的距离编号
q.push(v);
}
}
}
return dis[t] != -1; //确认是否连通
}
int dfs(int u, int flow_in)
{
if (u == t) return flow_in;
int flow_out = 0; //记录这一点实际流出的流量
for (int i = head[u]; i != -1;i = edge[i].next)
{
int v = edge[i].to;
if (dis[v] == dis[u] + 1 && edge[i].flow > 0)
{
int flow_part = dfs(v, min(flow_in, edge[i].flow));
if (flow_part == 0)continue; //无法形成增广路
flow_in -= flow_part; //流出了一部分，剩余可分配流入就减少了
flow_out += flow_part; //记录这一点最大的流出
edge[i].flow -= flow_part;
edge[i ^ 1].flow += flow_part; //减少增广路上边的容量，增加其反向边的容量
if (flow_in == 0)
break;
}
}
return flow_out;
}
int max_flow()
{
int sum = 0;
while (bfs())
{
sum += dfs(s, inf);
}
return sum;
}
int main() {
#ifdef LOCAL
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
while (scanf("%d%d%d%d", &n, &m, &s, &t) != EOF)
{
init();
for (int i = 1, u, v, flow;i <= m; i++)
{
scanf("%d%d%d", &u, &v, &flow);
add(u, v, flow);add(v, u, 0);
}
printf("%d\n", max_flow());
}
return 0;
}
二分图匹配
要解决这类问题，我们需要先了解什么是二分图？
二分图：一个图中的所有顶点可以分为两个集合 V,K ,其实两个集合内部的点彼此之间无边，如下图所示：（蓝色的点和红色的点分属于两个集合V,K)
然后我们回到这个题目上来，这个题目求的是最大可出战人数，实际上就是在二分图中找到两个集合中的最大匹配数，这类问题我们称之为二分图最大匹配数问题
属于网络流经典题目之一，下面说明一下建图的过程
1）由源点向集合V中每个点建一条容量为1的边
2）对于V，K集合之间存在的边e，v 为V中的点,k为K中的点，我们建一条容量为1的边，方向为 v --> k
3）由K中每个点向汇点建一条容量为1的边
当我们将图建好了后，我们求这个图的最大流，这个最大流即为二分图最大匹配数，下面展示一下建成的图：(S代表源点，T代表汇点，蓝色的边代表容量为1的边）
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<string>
#include<fstream>
#include<vector>
#include<stack>
#include <map>
#include <iomanip>
#define bug cout << "**********" << endl
#define show(x,y) cout<<"["<<x<<","<<y<<"] "
//#define LOCAL = 1;
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll mod = 1e6 + 3;
const int Max = 1e6 + 10;
struct Edge
{
int to, next, flow;
}edge[Max << 1];;
int n, m, a, b, s, t;
int head[Max], tot;
int dis[Max];
int ans;
bool vis[Max];
void init()
{
memset(head, -1, sizeof(head));tot = 0;
ans = 0;
}
void add(int u, int v, int flow)
{
edge[tot].to = v;
edge[tot].flow = flow;
edge[tot].next = head[u];
head[u] = tot++;
}
bool bfs()
{
memset(dis, -1, sizeof(dis));
dis[s] = 0;
queue<int>q;
q.push(s);
while (!q.empty())
{
int u = q.front();q.pop();
for (int i = head[u]; i != -1;i = edge[i].next)
{
int v = edge[i].to;
if (dis[v] == -1 && edge[i].flow > 0)
{
dis[v] = dis[u] + 1;
if (v == t) return true;
q.push(v);
}
}
}
return false;
}
int dfs(int u, int flow_in)
{
if (u == t) return flow_in;
int flow_out = 0;
for (int i = head[u]; i != -1;i = edge[i].next)
{
int v = edge[i].to;
if (dis[v] == dis[u] + 1 && edge[i].flow > 0)
{
int flow_part = dfs(v, min(flow_in, edge[i].flow));
if (flow_part == 0) continue;
flow_in -= flow_part;
flow_out += flow_part;
edge[i].flow -= flow_part;
edge[i ^ 1].flow += flow_part;
if (flow_in == 0)break;
}
}
return flow_out;
}
int max_val()
{
int sum = 0;
while (bfs())
{
sum += dfs(s, inf);
}
return sum;
}
int main()
{
#ifdef LOCAL
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
while (scanf("%d%d", &m, &n) != EOF)
{
init();
s = 0, t = n + 1;
for (int i = 1;i <= m;i++)
{
add(s, i, 1);add(i, s, 0); //由源点向外籍飞行员建边
}
for (int i = m + 1; i <= n;i++)
{
add(i, t, 1);add(t, i, 0);
}
while (scanf("%d%d", &a, &b) != EOF && a != -1 && b != -1)
{
add(a, b, 1);add(b, a, 0);
}
printf("%d\n", max_val());
for (int u = 1;u <= m;u++)
{
for (int i = head[u]; i != -1;i = edge[i].next)
{
if (edge[i].flow == 0 && edge[i].to != s && edge[i].to != t)
{
printf("%d %d\n", u, edge[i].to);
}
}
}
}
return 0;
}
最小费用最大流
这类题目相比于最大流问题新增了每天边单位流量的价格，问在最大流的情况下求出最小的费用。
这类题目和最大流很想，不过也有不小区别，对于这类问题，我们为每条边建的反边的价格是每天边的相反数，如图
然后我们的算法也不再是Dinic算法了，而是用spfa或者dijkstra
#pragma GCC optimize(2)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<string>
#include<fstream>
#include<vector>
#include<stack>
#include <map>
#include <iomanip>
#define bug cout << "**********" << endl
#define show(x,y) cout<<"["<<x<<","<<y<<"] "
#define LOCAL = 1;
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll mod = 1e9 + 7;
const int Max = 5e3 + 10;
struct Edge
{
int to, rev; //rev记录反向边
int flow, cost;;
};
int n, m, k;
vector<Edge>edge[Max << 1];
int h[Max]; //每个结点的势
int dis[Max];
int pre_node[Max], pre_edge[Max]; //前驱结点和对应边
void add(int u, int v, int flow, int cost)
{
edge[u].push_back({ v,(int)edge[v].size(),flow,cost });
edge[v].push_back({ u,(int)edge[u].size() - 1,0,-cost });
}
void min_cost_flow(int s, int t, int& min_cost, int& max_flow)
{
fill(h + 1, h + 1 + n, 0);
min_cost = max_flow = 0;
int tot = inf; //源点流量无限
while (tot > 0)
{
priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > >q;
memset(dis, inf, sizeof(dis));
dis[s] = 0;q.push({ 0,s });
while (!q.empty())
{
int u = q.top().second;
int dist = q.top().first;
q.pop();
if (dis[u] < dist)continue; //当前的距离不是最近距离
for (int i = 0;i < edge[u].size(); i++)
{
Edge &e = edge[u][i];
if (edge[u][i].flow > 0 && dis[e.to] > dis[u] + e.cost + h[u] - h[e.to])
{
dis[e.to] = dis[u] +e.cost + h[u] - h[e.to];
pre_node[e.to] = u;
pre_edge[e.to] = i;
q.push({ dis[e.to],e.to });
}
}
}
if (dis[t] == inf)break; //无法增广了，就是找到答案了
for (int i = 1;i <= n;i++) h[i] += dis[i];
int flow = tot; //求这一增广路径的流量
for (int i = t; i != s; i = pre_node[i])
flow = min(flow, edge[pre_node[i]][pre_edge[i]].flow);
for (int i = t; i != s; i = pre_node[i])
{
Edge& e = edge[pre_node[i]][pre_edge[i]];
e.flow -= flow;
edge[i][e.rev].flow += flow;
}
tot -= flow;
max_flow += flow;
min_cost += flow * h[t];
}
}
int main()
{
#ifdef LOCAL
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#endif
int s, t;
while (scanf("%d%d%d%d", &n, &m, &s, &t) != EOF)
{
for (int i = 1, u, v, flow, cost;i <= m;i++)
{
scanf("%d%d%d%d", &u, &v, &flow, &cost);
add(u, v, flow, cost);
}
int min_cost, max_flow;
min_cost_flow(s, t, min_cost, max_flow);
printf("%d %d\n", max_flow, min_cost);
}
return 0;
}
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<string>
#include<fstream>
#include<vector>
#include<stack>
#include <map>
#include <iomanip>
#define bug cout << "**********" << endl
#define show(x,y) cout<<"["<<x<<","<<y<<"] "
//#define LOCAL = 1;
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll mod = 1e9 + 7;
const int Max = 1e5 + 10;
struct Edge
{
int to, next;
int flow, cost;
}edge[Max << 1];
int n, m, s, t;
int head[Max], tot;
int dis[Max];
int pre[Max]; //记录增广路径此点的前一天边
bool vis[Max];
void init()
{
memset(head, -1, sizeof(head));tot = 0;
}
void add(int u, int v, int flow, int cost)
{
edge[tot].to = v;
edge[tot].flow = flow;
edge[tot].cost = cost;
edge[tot].next = head[u];
head[u] = tot++;
edge[tot].to = u;
edge[tot].flow = 0;
edge[tot].cost = -cost;
edge[tot].next = head[v];
head[v] = tot++;
}
bool spfa(int s, int t)
{
memset(dis, inf, sizeof(dis));
memset(vis, 0, sizeof(vis));
memset(pre, -1, sizeof(pre));
queue<int>q;
q.push(s);dis[s] = 0;vis[s] = true;
while (!q.empty())
{
int u = q.front();q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].flow > 0 && dis[v] > dis[u] + edge[i].cost)
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if (!vis[v])
{
vis[v] = true;q.push(v);
}
}
}
}
return pre[t] != -1;
}
void min_cost_max_flow(int s, int t, int& max_flow, int& min_cost)
{
max_flow = 0;
min_cost = 0;
while (spfa(s, t))
{
int flow = inf;
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) //沿增广路回溯edge[i^1]即为其反边
{
flow = min(flow, edge[i].flow);
}
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
edge[i].flow -= flow;
edge[i ^ 1].flow += flow;
min_cost += flow * edge[i].cost;
}
max_flow += flow;
}
}
int main()
{
#ifdef LOCAL
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
while (scanf("%d%d%d%d", &n, &m, &s, &t) != EOF)
{
init();
for (int i = 1, u, v, flow, cost;i <= m;i++)
{
scanf("%d%d%d%d", &u, &v, &flow, &cost);
add(u, v, flow, cost);
}
int max_flow = 0, min_cost = 0;
min_cost_max_flow(s, t, max_flow, min_cost);
printf("%d %d\n", max_flow, min_cost);
}
return 0;
}