uva 11478 Halum（图论-差分约束）

 

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 InputSample 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

   

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Problem Setter: Md. Mahbubul Hasan
Next Generation Contest 5

   

题目大意：

你可以给每个点的入边加一个值和出边加一个值，问你最小的边权最大是多少？

解题思路：

用二分枚举答案假设为x，那么 w(a,b)+sum[a]-sum[b]>=x，这些不等式构成了差分约束系统。

解题代码：

#include <iostream>#include <cstdio>#include <vector>#include <queue>#include <cmath>using namespace std;const int maxn=510;const int maxm=6000;const int inf=100000;struct edge{    int u,v,w,next;}e[maxm];int head[maxn],cnt,n,m,maxr;int dis[maxn],num[maxn];bool visited[maxn];vector <edge> v;void input(){    maxr=-(1<<30);    v.clear();    v.resize(m);    for(int i=0;i<m;i++){        scanf("%d%d%d",&v[i].u,&v[i].v,&v[i].w);        if(v[i].w>maxr) maxr=v[i].w;    }}bool spfa(){    for(int i=0;i<=n;i++){        dis[i]=inf;        visited[i]=false;        num[i]=0;    }    queue <int> q;    q.push(0);    visited[0]=true;    dis[0]=0;    num[0]++;    while(!q.empty()){        int s=q.front();        q.pop();        visited[s]=false;        for(int i=head[s];i!=-1;i=e[i].next){            int d=e[i].v;            if(dis[d]>dis[s]+e[i].w){                dis[d]=dis[s]+e[i].w;                if(!visited[d]){                    visited[d]=true;                    q.push(d);                    if(++num[d]>(int)sqrt(n+1)+1 ) return false;                }            }        }    }    return true;}void adde(int u,int v,int w){    e[cnt].u=u,e[cnt].v=v,e[cnt].w=w,e[cnt].next=head[u],head[u]=cnt++;}bool can(int x){    cnt=0;    for(int i=0;i<=n;i++) head[i]=-1;    for(int i=1;i<=n;i++) adde(0,i,0);    for(int i=0;i<m;i++){        adde(v[i].u,v[i].v,v[i].w-x);    }    return spfa();}void solve(){    if(can(maxr+1)) {        printf("Infinite\n");        return;    }    if(!can(1)) {        printf("No Solution\n");        return;    }    int l=1,r=maxr+1;    while(l<r){        int mid=(l+r)/2;        if(!can(mid)) r=mid;        else l=mid+1;    }    printf("%d\n",r-1);}int main(){    while(scanf("%d%d",&n,&m)!=EOF){        input();        solve();    }    return 0;}