标签：style blog color os io for 2014 ar
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Problem H
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Halum
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Time Limit : 3 seconds
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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.
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| Input | ||||||
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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. |
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| Output | ||||||
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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” |
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| Sample Input | Sample Output | |||||
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2 1
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Infinite |
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Problem Setter: Md. Mahbubul Hasan |
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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：
解题代码：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。
<pre name="code" class="cpp">#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
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]>n+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;
}
uva 11478 Halum（图论-差分约束）,布布扣,bubuko.com
标签：style blog color os io for 2014 ar
原文地址：http://blog.csdn.net/a1061747415/article/details/38436295
