UVA 10594 Data Flow

无向图费用流

还有一段话摘自别人博客

这道题是无向图的最小费用最大流问题，看清楚是无向图的。这么说无向图和有向图的费用流问题有什么区别呢？主要是反向边的问题。首先我们说一下最大流问题中的反向边，我们需要将其cap[u][v]=0表示容量为0，而在费用流问题中添加了费用，所以肯定不能像之前那么简单处理了，那怎么办呢？在有向图中，没有存在的反向边我们用cap[u][v]=0表示容量为0，cost[v][u]=-cost[u][v]表示取反的费用，简单说就是讲这部分费用减除，相当于没有走。 现在可以说一下无向图和有向图的不同了，既然两个方向都是可以走的，那么我们就将原本有的一条边变化出了四条边，两个原有边，两个反向边，原有两个边相互独立，不能将这两个原有边看成互为反向边，否则就出现了环路，spfa就走不通

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
const LL INF = 100000000;
#define MAXN 110
struct node
{
    int u,v,next;
    LL cap,flow,cost;
}edge[5005 * 4];
int N,M,cnt,src,tag;
LL K,D,F,C,d[5005];
struct point
{
    LL x,y,w;
}res[5005];
bool inq[MAXN];
int head[MAXN],p[MAXN];
void add(int u, int v, LL cost, LL cap)
{
    edge[cnt].v = v;
    edge[cnt].u = u;
    edge[cnt].cost = cost;
    edge[cnt].cap = cap;
    edge[cnt].flow = 0;
    edge[cnt].next = head[u];
    head[u] = cnt++;
    // 反向边
    edge[cnt].v = u;
    edge[cnt].u = v;
    edge[cnt].cost = -cost;
    edge[cnt].cap = 0;
    edge[cnt].flow = 0;
    edge[cnt].next = head[v];
    head[v] = cnt++;
}
void read()
{
    for (int i = 1; i <= M; i++) scanf("%lld%lld%lld",&res[i].x,&res[i].y,&res[i].w);
    scanf("%lld%lld",&D,&K);
    cnt = 0;
    src = 0;
    tag = N;
    memset(head,-1,sizeof(head));
    for (int i = 1; i <= M; i++)
    {
       add(res[i].x,res[i].y,res[i].w,K);
       add(res[i].y,res[i].x,res[i].w,K);
    }
    add(0,1,0,D);
}
bool SPFA()
{
    queue<int>q; while (!q.empty()) q.pop();
    for (int i = 0; i < MAXN; i++) d[i] = INF;
    d[src] = 0;
    memset(p,-1,sizeof(p));
    memset(inq,false,sizeof(inq));
    q.push(src);
    inq[src] = true;
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        inq[u] = false;
        for (int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            if (edge[i].cap > edge[i].flow && d[v] > d[u] + edge[i].cost)
            {
                d[v] = d[u] + edge[i].cost;
                p[v] = i;
                if (!inq[v])
                {
                   inq[v] = true;
                   q.push(v);
                }
            }
        }
    }
    //printf("%lld
",d[tag]);
    return   d[tag] != INF;
}
void slove()
{
    F = C = 0;
    while (SPFA())
    {
        LL a = INF;
        for (int i = p[tag]; i != -1; i = p[edge[i].u])
             a = min(a,edge[i].cap - edge[i].flow);
        for (int i = p[tag]; i != -1; i = p[edge[i].u])
        {
            edge[i].flow += a;
            edge[i ^ 1].flow -= a;
        }
        F += a;
        C += d[tag] * a;
    }
}
int main()
{
    //freopen("sample.txt","r",stdin);
    while (scanf("%d%d",&N,&M) != EOF)
    {
        read();
        slove();
        if (F == D) printf("%lld
",C);
        else puts("Impossible.");
    }
    return 0;
}