POJ 2396 Budget 上下界网络流 - c++编程基础

这是一道比较基础的上下界网络流了
上下界网络流的算法完全是参考论文《一种简易的方法求解流量有上下界的网络》还有《最大流在信息学竞赛中应用的一个模型》
然后说的也是比较详细
具体的解法不再多说了，网络流的算法一直是比较麻烦，代码量还大，我看了好久才明白算法是啥意思。
刚开始比较迷惑为啥加了一对源汇了，又加一对。后来明白了，对本题而言，刚开始是一个有源汇的网络，
就是我们加的第一对的源汇，而我们要转化为一个无源汇的网络，就在汇到源加一条无穷容量的边，这样就满足了定义的要求，后来再加一对的源汇，才是论文中说的附加源汇
代码比较肥硕
输出答案的时候。刚开始觉得挺蛋疼，后来发现边都是按顺序加的，瞬间觉得世界又美好了
我的模板里，边里存的cap表示这条边还有多少流量可用，flow就是现在已经使用的流量

[cpp]
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define MAXN 555
#define MAXM 555555
#define INF 1000000007
using namespace std;
struct node
{
int ver; // vertex
int cap; // capacity
int flow; // current flow in this arc
int next, rev;
}edge[MAXM];
int dist[MAXN], numbs[MAXN], src, des, n;
int head[MAXN], e;
void add(int x, int y, int c)
{ //e记录边的总数
edge[e].ver = y;
edge[e].cap = c;
edge[e].flow = 0;
edge[e].rev = e + 1; //反向边在edge中的下标位置
edge[e].next = head[x]; //记录以x为起点的上一条边在edge中的下标位置
head[x] = e++; //以x为起点的边的位置
//反向边
edge[e].ver = x;
edge[e].cap = 0; //反向边的初始可行流量为0
edge[e].flow = 0;
edge[e].rev = e - 1;
edge[e].next = head[y];
head[y] = e++;
}
void rev_BFS()
{
int Q[MAXN], qhead = 0, qtail = 0;
for(int i = 1; i <= n; ++i)
{
dist[i] = MAXN;
numbs[i] = 0;
}
Q[qtail++] = des;
dist[des] = 0;
numbs[0] = 1;

while(qhead != qtail)
{
int v = Q[qhead++];
for(int i = head[v]; i != -1; i = edge[i].next)
{
if(edge[edge[i].rev].cap == 0 || dist[edge[i].ver] < MAXN)continue;
dist[edge[i].ver] = dist[v] + 1;
++numbs[dist[edge[i].ver]];
Q[qtail++] = edge[i].ver;
}
}
}
void init()
{
e = 0;
memset(head, -1, sizeof(head));
}
int maxflow()
{
int u, totalflow = 0;
int Curhead[MAXN], revpath[MAXN];
for(int i = 1; i <= n; ++i)Curhead[i] = head[i];
u = src;
while(dist[src] < n)
{
if(u == des) // find an augmenting path
{
int augflow = INF;
for(int i = src; i != des; i = edge[Curhead[i]].ver)
augflow = min(augflow, edge[Curhead[i]].cap);
for(int i = src; i != des; i = edge[Curhead[i]].ver)
{
edge[Curhead[i]].cap -= augflow;
edge[edge[Curhead[i]].rev].cap += augflow;
edge[Curhead[i]].flow += augflow;
edge[edge[Curhead[i]].rev].flow -= augflow;
}
totalflow += augflow;
u = src;
}
int i;
for(i = Curhead[u]; i != -1; i = edge[i].next)
if(edge[i].cap > 0 && dist[u] == dist[edge[i].ver] + 1)break;
if(i != -1) // find an admissible arc, then Advance
{
Curhead[u] = i;
revpath[edge[i].ver] = edge[i].rev;
u = edge[i].ver;
}
else // no admissible arc, then relabel this vertex
{
if(0 == (--numbs[dist[u]]))break; // GAP cut, Important!
Curhead[u] = head[u];
int mindist = n;
for(int j = head[u]; j != -1; j = edge[j].next)
if(edge[j].cap > 0)mindist = min(mindist, dist[edge[j].ver]);
dist[u] = mindist + 1;
++numbs[dist[u]];

POJ 2396 Budget 上下界网络流(一)