| Language:Default Ombrophobic Bovines
Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to create a rain evacuation plan so that all the cows can get to shelter before the rain begins. Weather forecasting is not always correct, though. In order to minimize false alarms, they want to sound the siren as late as possible while still giving enough time for all the cows to get to some shelter. Input * Line 1: Two space-separated integers: F and P Output * Line 1: The minimum amount of time required for all cows to get under a shelter, presuming they plan their routes optimally. If it not possible for the all the cows to get under a shelter, output "-1". Sample Input Sample Output Hint OUTPUT DETAILS: Source |
给定一个有n个顶点和m条边的无向图，点i 处有Ai头牛，点i 处的牛棚能容纳Bi头牛，每条边有一个时间花费ti(表示从一个端点走到另一个端点所需要的时间)，求一个最短时间T使得在T时间内所有的牛都能进到某一牛棚里去。
将每个点i 拆成两个点i’, i’’，连 边(s, i’, Ai), (i’’, t, Bi)。二分最短时间T，若d[i][j]<=T（d[i][j]表示点i, j 之间的最短时间花费）则加边(i’, j’’, ∞)。每次根据最大流调整二分的上下界即可。
这里拆点是因为i点有两个属性，一个是现在有多少牛，其次是一共能容纳多少牛，加边时i和i''之间加无穷，因为一个棚可以放多只奶牛
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#define inf 1e9
#define infLL 1LL<<60
using namespace std;
const int maxn=500+10;
struct Edge
{
int from,to,cap,flow;
Edge(){}
Edge(int f,int t,int c,int fl):from(f),to(t),cap(c),flow(fl){}
};
struct Dinic
{
int n,m,s,t;
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn];
bool vis[maxn];
int cur[maxn];
void init(int n,int s,int t)
{
this->n=n,this->s=s,this->t=t;
edges.clear();
for(int i=0;i<n;i++) G[i].clear();
}
void addedge(int from,int to,int cap)
{
edges.push_back( Edge(from,to,cap,0) );
edges.push_back( Edge(to,from,0,0) );
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool BFS()
{
queue<int> Q;
memset(vis,0,sizeof(vis));
vis[s]=true;
d[s]=0;
Q.push(s);
while(!Q.empty())
{
int x=Q.front(); Q.pop();
for(int i=0;i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(!vis[e.to] && e.cap>e.flow)
{
vis[e.to]=true;
d[e.to]=d[x]+1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x,int a)
{
if(x==t || a==0) return a;
int flow=0,f;
for(int& i=cur[x];i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(d[e.to]==d[x]+1 && (f=DFS(e.to,min(a,e.cap-e.flow) ) ) >0)
{
e.flow+=f;
edges[G[x][i]^1].flow -=f;
flow+=f;
a-=f;
if(a==0) break;
}
}
return flow;
}
int maxflow()
{
int ans=0;
while(BFS())
{
memset(cur,0,sizeof(cur));
ans +=DFS(s,inf);
}
return ans;
}
}Dc;
int n,m;
long long dis[maxn][maxn];
int now[maxn];
int can[maxn];
int fullflow;
void floyd(int n)
{
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(dis[i][k]<infLL&&dis[k][j]<infLL)
dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
}
bool solve(long long limit)
{
Dc.init(2*n+2,0,2*n+1);
for(int i=1;i<=n;i++)
{
Dc.addedge(0,i,now[i]);
Dc.addedge(i+n,2*n+1,can[i]);
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(dis[i][j]<=limit)
Dc.addedge(i,j+n,inf);
return Dc.maxflow()==fullflow;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{fullflow=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
dis[i][j]=i==j?0:infLL;
for(int i=1;i<=n;i++)
{scanf("%d%d",&now[i],&can[i]);
fullflow+=now[i];
}
int u,v;
long long w;
while(m--)
{
scanf("%d%d%lld",&u,&v,&w);
dis[u][v]=dis[v][u]=min(w,dis[u][v]);
}
floyd(n);
long long l=0,r=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(dis[i][j]<infLL)
r=max(r,dis[i][j]);
if(!solve(r))
printf("-1\n");
else
{
while(r>l)
{
long long mid=l+(r-l)/2;
if(solve(mid))
r=mid;
else
l=mid+1;
}
printf("%lld\n",r);
}
}
return 0;
}