POJ 2391 二分+最大流

Ombrophobic Bovines

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19066		Accepted: 4138

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. 

The farm has F (1 <= F <= 200) fields on which the cows graze. A set of P (1 <= P <= 1500) paths connects them. The paths are wide, so that any number of cows can traverse a path in either direction. 

Some of the farm's fields have rain shelters under which the cows can shield themselves. These shelters are of limited size, so a single shelter might not be able to hold all the cows. Fields are small compared to the paths and require no time for cows to traverse. 

Compute the minimum amount of time before rain starts that the siren must be sounded so that every cow can get to some shelter.

Input

* Line 1: Two space-separated integers: F and P 

* Lines 2..F+1: Two space-separated integers that describe a field. The first integer (range: 0..1000) is the number of cows in that field. The second integer (range: 0..1000) is the number of cows the shelter in that field can hold. Line i+1 describes field i. 

* Lines F+2..F+P+1: Three space-separated integers that describe a path. The first and second integers (both range 1..F) tell the fields connected by the path. The third integer (range: 1..1,000,000,000) is how long any cow takes to traverse it.

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

3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120

Sample Output

110

Hint

OUTPUT DETAILS: 

In 110 time units, two cows from field 1 can get under the shelter in that field, four cows from field 1 can get under the shelter in field 2, and one cow can get to field 3 and join the cows from that field under the shelter in field 3. Although there are other plans that will get all the cows under a shelter, none will do it in fewer than 110 time units.

Source

USACO 2005 March Gold

题意：

有n块地，每块地初始有a头牛和一个能容纳b头牛的牛棚，有m条带权边，表示走过这条边需要的时间

，问最少需要多少时间能让所有的牛在进入棚子（n头牛同时行动）。

代码：

//先floyd求出每两个点之间的最短距离,二分距离，如果两点之间的距离小于等于此
//二分值mid，两点之间建容量为inf的边，还要拆点，源点到i建容量为i点牛数量的
//边，i+n到汇点建容量为i点牛棚能容纳牛数量的边，看最大流是否是牛的总数。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn=410;
const int inf=0x7fffffff;
const ll llinf=(1LL<<60);
struct Edge{
    int from,to,cap,flow;
    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct Dinic{
    int n,m,s,t;
    vector<Edge>edges;
    vector<int>g[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    void Init(int n){
        this->n=n;
        for(int i=0;i<n;i++) g[i].clear();
        edges.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));//反向弧
        m=edges.size();
        g[from].push_back(m-2);
        g[to].push_back(m-1);
    }
    bool Bfs(){
        memset(vis,0,sizeof(vis));
        queue<int>q;
        q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!q.empty()){
            int x=q.front();q.pop();
            for(int i=0;i<(int)g[x].size();i++){
                Edge &e=edges[g[x][i]];
                if(!vis[e.to]&&e.cap>e.flow){
                    vis[e.to]=1;
                    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<(int)g[x].size();i++){
            Edge &e=edges[g[x][i]];
            if(d[x]+1==d[e.to]&&(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 s,int t){
        this->s=s;this->t=t;
        int flow=0;
        while(Bfs()){
            memset(cur,0,sizeof(cur));
            flow+=Dfs(s,inf);
        }
        return flow;
    }
}dc;
int N,M,a,b,c,x[maxn],y[maxn];
ll mp[maxn][maxn];
void Floyd(ll &r){
    for(int k=1;k<=N;k++)
        for(int i=1;i<=N;i++)
            for(int j=0;j<=N;j++)
                if(mp[i][k]!=llinf&&mp[k][j]!=llinf)
                mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
    for(int i=1;i<=N;i++)
        for(int j=1;j<=N;j++){
            if(mp[i][j]!=llinf) r=max(r,mp[i][j]);
        }
}
int Solve(ll mid){
    dc.Init(2*N+2);
    for(int i=1;i<=N;i++){
        dc.Addedge(0,i,x[i]);
        dc.Addedge(i+N,2*N+1,y[i]);
        for(int j=1;j<=N;j++)
            if(mp[i][j]<=mid)
                dc.Addedge(i,j+N,inf);
    }
    return dc.Maxflow(0,2*N+1);
}
int main()
{
    while(scanf("%d%d",&N,&M)==2){
        int sum1=0,sum2=0;
        for(int i=1;i<=N;i++){
            scanf("%d%d",&x[i],&y[i]);
            sum1+=x[i];sum2+=y[i];
        }
        for(int i=1;i<=N;i++)
            for(int j=1;j<=N;j++)
                mp[i][j]=(i==j?0:llinf);
        while(M--){
            scanf("%d%d%d",&a,&b,&c);
            mp[a][b]=mp[b][a]=min(mp[a][b],1LL*c);
        }
        ll l=0,r=0,mid,ans;
        Floyd(r);
        if(Solve(r)!=sum1){
            printf("-1
");
            continue;
        }
        while(l<=r){
            mid=(l+r)>>1;
            if(Solve(mid)==sum1){
                ans=mid;r=mid-1;
            }
            else l=mid+1;
        }
        printf("%I64d
",ans);
    }
    return 0;
}