HDU3157 Crazy Circuits

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有源汇的上下界网络流求最小流。

这个其实和上道题差不多……题目描述我没怎么看明白……好像就是让你按照他说的把图建出来就行了，注意这个题的字符处理，可能有长度大于1的字符串，要注意一下。求最小流的话还是先求可行流，之后因为可行流可能流多，而从汇点向原点跑相当于退流。所以我们再从原点向汇点跑一次最大流，两次结果相减就是答案。

然后这个题其实挺奇怪的……辅助源汇点连接的边其实删不删无所谓，HDU的数据不知道行不行……然而POJ这题没数据，交个空程序就能过……

看一下代码。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<vector>
#include<map>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar(‘\n‘)
#define fr friend inline
#define y1 poj
#define mp make_pair
#define pr pair<int,int>
#define fi first
#define sc second
#define pb push_back

using namespace std;
typedef long long ll;
const int M = 40005;
const int N = 100005;
const int INF = 0x3f3f3f3f;
const double eps = 1e-7;

int read()
{
    int ans = 0,op = 1;char ch = getchar();
    while(ch < ‘0‘ || ch > ‘9‘) {if(ch == ‘-‘) op = -1;ch = getchar();}
    while(ch >= ‘0‘ && ch <= ‘9‘) ans = ans * 10 + ch - ‘0‘,ch = getchar();
    return ans * op;
}

struct edge
{
   int next,to,from,v;
}e[N<<3];

int head[N],cur[N],deg[N],x,y,z,n,m,g,ecnt,S,T,S1,T1,c,d,tot;
int low[2005][2005],dep[N],sta[N],top;
queue <int> q;
char s[3];

void add(int x,int y,int z)
{
   e[++ecnt].to = y;
   e[ecnt].next = head[x];
   e[ecnt].v = z;
   head[x] = ecnt;
}

void clear()
{
   memset(head,-1,sizeof(head)),ecnt = -1,tot = 0;
   memset(deg,0,sizeof(deg));
}

bool bfs(int s,int t)
{
   while(!q.empty()) q.pop();
   rep(i,0,t) cur[i] = head[i];
   memset(dep,-1,sizeof(dep));
   dep[s] = 0,q.push(s);
   while(!q.empty())
   {
      int k = q.front();q.pop();
      for(int i = head[k];~i;i = e[i].next)
      {
     if(e[i].v && dep[e[i].to] == -1)
        dep[e[i].to] = dep[k] + 1,q.push(e[i].to);
      }
   }
   return dep[t] != -1;
}

int dfs(int s,int t,int lim)
{
   if(s == t || !lim) return lim;
   int flow = 0;
   for(int i = cur[s];~i;i = e[i].next)
   {
      cur[s] = i;
      if(dep[e[i].to] != dep[s] + 1) continue;
      int f = dfs(e[i].to,t,min(lim,e[i].v));
      if(f)
      {
     e[i].v -= f,e[i^1].v += f;
     flow += f,lim -= f;
     if(!lim) break;
      }
   }
   if(!flow) dep[s] = -1;
   return flow;
}

int dinic(int s,int t)
{
   int maxflow = 0;
   while(bfs(s,t)) maxflow += dfs(s,t,INF);
   return maxflow;
}

int change(char *c)
{
   if(c[0] == ‘+‘) return S;
   if(c[0] == ‘-‘) return T;
   int l = strlen(c),cur = 0;
   rep(i,0,l-1) cur *= 10,cur += c[i] - ‘0‘;
   return cur;
}     

void rebuild()
{
   e[ecnt].v = e[ecnt-1].v = 0;
   //for(int i = head[S1];~i;i = e[i].next) e[i].v = e[i^1].v = 0;
   //for(int i = head[T1];~i;i = e[i].next) e[i].v = e[i^1].v = 0;
}

int main()
{
   //freopen("f.in","r",stdin);
   //freopen("f.out","w",stdout);
   while(scanf("%d%d",&n,&m))
   {
      if(!n && !m) break;
      T = n + 1,S1 = T + 1,T1 = S1 + 1;
      clear();
      rep(i,1,m)
      {
     scanf("%s",s),x = change(s);
     scanf("%s",s),y = change(s);
     z = read(),add(x,y,INF-z),add(y,x,0),deg[x] += z,deg[y] -= z;
      }
      rep(i,S,T)
      {
     if(deg[i] > 0) add(i,T1,deg[i]),add(T1,i,0),tot += deg[i];
     else add(S1,i,-deg[i]),add(i,S1,0);
      }
      add(T,S,INF),add(S,T,0);
      int g = dinic(S1,T1);
      if(g != tot) {printf("impossible\n");continue;}
      g = e[ecnt].v,rebuild();
      printf("%d\n",g - dinic(T,S));
   }
   return 0;
}

原文地址：https://www.cnblogs.com/captain1/p/10134839.html