UVA 11478V Halum 二分答案+差分约束系统

详细翻译版请见白皮书334页

11478 Halum

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

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

Sample Output

Infinite

Infinite

3

1


分析：

对于一个点的多次操作相对独立，可以合成一个操作，记为s[x];

题目要求的是最小值的最大值，容易想到二分答案（赤裸裸的暗示啊）。

二分答案最小值x，那么对于每一条边e 一定有：

e.w+s[e.from]-s[e.to]>=x   <==>    s[e.to]-s[e.from]<=e.w-x  ;

其实就是每条边的权重都减小x；

剩下的就是开心的差分啦~~

貌似比较简单，但是wa了6次，主要是因为判定负权回路的条件是进队次数达到n+1次（而不是n次）！！

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<cstdlib>

#define LL long long
#define CLEAR(XXX) memset((XXX),0,sizeof(XXX))
using namespace std;

const int maxn=500+5,maxm=5000+5,inf=1e9;

int m,n;
inline void _read(int &x){
char ch=getchar(); bool mark=false;
for(;!isdigit(ch);ch=getchar())if(ch=='-')mark=true;
for(x=0;isdigit(ch);ch=getchar())x=x*10+ch-'0';
if(mark)x=-x;
}
struct Edge{
int from,to,w;
Edge(int from,int to,int w):from(from),to(to),w(w){}
};
struct SPFA{
int n,m;
vector<Edge> edge;
int last[maxm],next[maxm];
LL dist[maxn];
int cnt[maxn];
bool vis[maxn];
void init(int n){
this->n = n;
m=0;
CLEAR(last); CLEAR(next);
edge.clear();
edge.push_back(Edge(0,0,0));
}
void add_edge(int from,int to,int dist){
edge.push_back(Edge(from,to,dist));
m=edge.size()-1;
next[m]=last[from];
last[from]=m;
}
bool solve(int s,int p){
int i;
CLEAR(vis); CLEAR(cnt);
for(i=1;i<=n;i++) dist[i]=inf;
dist[s]=0;
vis[s]=true;cnt[s]++;
queue <int> q;
q.push(s);
while(!q.empty()){
int x=q.front();
q.pop();vis[x]=false; //及时修改标记
for(i=last[x];i;i=next[i]){
Edge e=edge[i];
e.w-=p;
if(dist[e.from]+e.w<dist[e.to]){
dist[e.to]=dist[e.from]+e.w;
if(!vis[e.to]){
cnt[e.to]++; //统计入队次数，判断负权回路 注意是n+1次！！
if(cnt[e.to]==n+1)return false;
q.push(e.to) ;
vis[e.to]=true;
}
}
}
}
return true;
}
void answer(){
for(int i=1;i<=n;i++)
if(dist[i]>=inf)printf("NoPath\n");
else printf("%I64d\n",dist[i]);
}
};
SPFA solver;
int main(){
int i,x,y,d,l,r,ans=0;
while(cin>>n>>m){
solver.init(n);
for(i=1;i<=m;i++){
_read(x);_read(y);_read(d);
solver.add_edge(x,y,d);
}
for(i=1;i<=n;i++)solver.add_edge(0,i,0);
l=1;r=10001;
if(!solver.solve(0,1)){
puts("No Solution");continue;
}
if(solver.solve(0,r)) {
puts("Infinite");continue;
}
while(l<=r){
int mid=(l+r)>>1;
//w[x]+s[from]-s[to]>=mid; <=> s[to]-s[from]<=w[x]-mid;
if(solver.solve(0,mid)) l=mid+1,ans=max(ans,l);
else r=mid-1;
}
printf("%d\n",r);
}
return 0;
}