(网络流_最大流)Flow Problem

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 25;
typedef long long LL;
const int INF = 0xfffffff;
struct edge
{
int to, cap, rev;
edge(int to, int cap, int rev) :to(to), cap(cap), rev(rev){}
};

vector<edge> G[maxn];
bool used[maxn];
void add_edge(int from, int to, int cap)
{
G[from].push_back(edge( to, cap, G[to].size() ));
G[to].push_back(edge( from, 0, G[from].size() - 1 ));
}
int dfs(int v, int t, int f)
{
if (v == t) return f;
used[v] = true;
for (int i = 0; i < G[v].size(); i++)
{
edge &e = G[v][i];
if (!used[e.to] && e.cap>0)
{
int d = dfs(e.to, t, min(f, e.cap));
if (d > 0)
{
e.cap -= d;
G[e.to][e.rev].cap += d;
return d;
}
}
}
return 0;
}
int max_flow(int s, int t)
{
int flow = 0;
for (;;)
{
memset(used, 0, sizeof(used));
int f = dfs(s, t, INF);
if (f == 0) return flow;
flow += f;
}
}
int main()
{
//freopen("f:\\input.txt", "r", stdin);
int T;
int cas = 0;
scanf("%d", &T);
while (T--)
{
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)	G[i].clear();
for (int i = 0; i < m; i++)
{
int from, to, cap;
scanf("%d%d%d", &from, &to, &cap);
add_edge(from, to, cap);
}
printf("Case %d: %d\n", ++cas, max_flow(1, n));
}
return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 25;
typedef long long LL;
const int INF = 0xfffffff;
////////////////////////最大流开始//////////////////////////////////////
typedef int cap_type;
#define MAX_V 20

// 用于表示边的结构体（终点、容量、反向边）
struct edge
{
int to, rev;
cap_type cap;

edge(int to, cap_type cap, int rev) : to(to), cap(cap), rev(rev)
{}
};

vector <edge> G[MAX_V];   // 图的邻接表表示
int level[MAX_V];      // 顶点到源点的距离标号
int iter[MAX_V];       // 当前弧，在其之前的边已经没有用了

// 向图中加入一条从from到to的容量为cap的边
void add_edge(int from, int to, int cap)
{
G[from].push_back(edge(to, cap, G[to].size()));
G[to].push_back(edge(from, 0, G[from].size() - 1));
}

// 通过BFS计算从源点出发的距离标号
void bfs(int s)
{
memset(level, -1, sizeof(level));
queue<int> que;
level[s] = 0;
que.push(s);
while (!que.empty())
{
int v = que.front();
que.pop();
for (int i = 0; i < G[v].size(); ++i)
{
edge &e = G[v][i];
if (e.cap > 0 && level[e.to] < 0)
{
level[e.to] = level[v] + 1;
que.push(e.to);
}
}
}
}

// 通过DFS寻找增广路
cap_type dfs(int v, int t, cap_type f)
{
if (v == t)
{
return f;
}
for (int &i = iter[v]; i < G[v].size(); ++i)
{
edge &e = G[v][i];
if (e.cap > 0 && level[v] < level[e.to])
{
cap_type d = dfs(e.to, t, min(f, e.cap));
if (d > 0)
{
e.cap -= d;
G[e.to][e.rev].cap += d;
return d;
}
}
}

return 0;
}

// 求解从s到t的最大流
cap_type max_flow(int s, int t)
{
cap_type flow = 0;
for (;;)
{
bfs(s);
if (level[t] < 0)
{
return flow;
}
memset(iter, 0, sizeof(iter));
cap_type f;
while ((f = dfs(s, t, 0x3f3f3f3f3f3f3f3f)) > 0)
{
flow += f;
}
}
}

///////////////////////////////最大流结束/////////////////////////////////////
int main()
{
//freopen("f:\\input.txt", "r", stdin);
int T;
int cas = 0;
scanf("%d", &T);
while (T--)
{
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)	G[i].clear();
for (int i = 0; i < m; i++)
{
int from, to, cap;
scanf("%d%d%d", &from, &to, &cap);
add_edge(from, to, cap);
}
printf("Case %d: %d\n", ++cas, max_flow(1, n));
}
return 0;
}