UVA 11478 Halum(差分约束，5级)

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本文探讨了Halum操作在图算法中的应用，通过调整边的成本来使所有边成本达到正数的最大值。介绍了如何使用差分约束系统解决此问题，并提供了一段实现该算法的C++代码。

Problem H	Halum	Time Limit : 3 seconds
You are given a directed graph G(V,E) with a set of vertices and edges. Each edge (i,j) that connects some vertex i to vertex j has an integer cost associated with that edge. Define the operation Halum(v, d) to operate on a vertex v using an integer d as follows: subtract d from the cost of all edges that enter v and add d to the cost of every edge that leaves v. As an example of that operation, consider graph G that has three vertices named (1, 2, 3) and two edges. Edge (1, 2) has cost -1, and edge (2,3) has cost 1. The operation Halum(2,-3) operates on edges entering and leaving vertex 2. Thus, edge (1, 2) gets cost -1-(-3)=2 and the edge (2, 3) gets cost 1 + (-3) = -2. Your goal is to apply the Halum function to a graph, potentially repeatedly, until every edge in the graph has at least a certain cost that is greater than zero. You have to maximize this cost.
Input
Two space-separated integers per case: V(V≤500) and E(E≤2700). E lines follow. Each line represents a directed edge using three space-separated integers (u, v, d). Absolute value of cost can be at most 10000.

Output
If the problem is solvable, then print the maximum possible value. If there is no such solution print “No Solution”. If the value can be arbitrary large print “Infinite”

Sample Input	Sample Output
2 1 1 2 10 2 1 1 2 -10 3 3 1 2 4 2 3 2 3 1 5 4 5 2 3 4 4 2 5 3 4 2 3 1 0 1 2 -1	Infinite Infinite 3 1
Problem Setter: Md. Mahbubul Hasan Next Generation Contest 5

思路：设sum(u)是u点所有加减操作的和，那么u->v的最终权值就是sum(u)+w(u,v)-sum(v)

二分最小权值x

那么就是 sum(u)+w(u,v)-sum(v)>=x

sum(v)<=sum(u)+w(u,v)-x 等价于最短路的d[v]<=d[u]+w(u,v);

因此一个差分就可以解决了。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int mm=500+9;
const int mn=4700+9;
const int oo=1e9;
int n,m,edge;
int head[mm];
class node
{
  public:int v,next,dis;
}e[mn];
void data()
{
 memset(head,-1,sizeof(head));edge=0;
}
void add(int u,int v,int _dis)
{
  e[edge].v=v;e[edge].dis=_dis;e[edge].next=head[u];head[u]=edge++;
}
int id[mm],dis[mm];bool vis[mm];
bool spfa(int x)
{
  int u,v;
  memset(vis,0,sizeof(vis));
  memset(id,0,sizeof(id));
  queue<int>Q;
  for(int i=0;i<=n;++i)
    dis[i]=oo;
  Q.push(0);vis[0]=1;
  dis[0]=0;
  int z,spend;
  while(!Q.empty())
  {
    u=Q.front();Q.pop();vis[u]=0;
    for(int i=head[u];~i;i=e[i].next)
    {
      v=e[i].v;
      spend=e[i].dis-x;
      if(dis[v]>dis[u]+spend)
      {
        dis[v]=dis[u]+spend;
        if(!vis[v])
        {
          if(++id[v]>=n+1)
          {return 0;}
          vis[v]=1;Q.push(v);
        }
      }
    }
  }
  return 1;
}
int main()
{ int a,b,c;
  while(~scanf("%d%d",&n,&m))
  {  data();
    for(int i=0;i<m;++i)
    {
      scanf("%d%d%d",&a,&b,&c);
      add(a,b,c);
    }
    for(int i=1;i<=n;++i)
      add(0,i,0);
    int l=0,r=10001,mid,ans=0;
    if(spfa(r)){printf("Infinite\n");continue;}
    if(!spfa(1)){printf("No Solution\n");continue;}
    while(l<=r)
    {
      mid=(l+r)/2;
      if(spfa(mid)){l=mid+1;ans=max(ans,mid);}
      else r=mid-1;
    }
    printf("%d\n",ans);
  }
  return 0;
}