poj 2391（二分+floyd+sap）

不难的一道网络流，因为为递增性，所以可以用二分，用二分还是相当的快，然后就是基本的建图，和sap了。

 

Ombrophobic Bovines

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10819		Accepted: 2419

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

#include <stdio.h>
#include <string.h>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
#define INF 0x3ffffff
#define MAX 1000000000000000LL
#define N 440
#define M 10000000
struct node
{
    int to,w,next;
}edge[M];

struct node1
{
    int u,v;
    __int64 w;
}g[1505];

int n,m;
int cnt,pre[N];
int s,t;
int savex[N],savey[N];
__int64 map[N][N];
int ans;
int lv[N],gap[N];
int nn;

void add_edge(int u,int v,__int64 w)
{
    edge[cnt].to=v;
    edge[cnt].w=w;
    edge[cnt].next=pre[u];
    pre[u]=cnt++;
}

int sdfs(int k,int w)
{
    if(k==t)
        return w;
    int f=0;
    int mi=nn-1;
    for(int p=pre[k];p!=-1;p=edge[p].next)
    {
        int v=edge[p].to;
        if(edge[p].w!=0)
        {
            if(lv[k]==lv[v]+1)
            {
                int tmp=sdfs(v,min(w-f,edge[p].w));
                f+=tmp;
                edge[p].w-=tmp;
                edge[p^1].w+=tmp;
                if(f==w||lv[s]==nn) return f;
            }
            if(lv[v]<mi) mi=lv[v];
        }
    }
    if(f==0)
    {
        gap[lv[k]]--;
        if( gap[lv[k]]==0 )
        {
            lv[s]=nn;
            return f;
        }
        lv[k]=mi+1;
        gap[lv[k]]++;
    }
    return f;
}

int sap()
{
    nn=2*n+2;
    int sum=0;
    memset(lv,0,sizeof(lv));
    memset(gap,0,sizeof(gap));
    gap[0]=nn;
    while(lv[s]<nn)
    {
        sum+=sdfs(s,INF);
    }
    return sum;
}

int check(__int64 mid)
{
    cnt=0;
    memset(pre,-1,sizeof(pre));
    for(int i=1;i<=n;i++)
    {    
        add_edge(s,i,savex[i]);
        add_edge(i,s,0);
        
        add_edge(n+i,t,savey[i]);
        add_edge(t,n+i,0);
        
        add_edge(i,n+i,INF);
        add_edge(n+i,i,0);
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(map[i][j]<=mid && map[i][j]!=0&&map[i][j]<MAX&&i!=j) //
            {
                add_edge(i,n+j,INF);
                add_edge(n+j,i,0);
            }
        }
    if(sap()==ans)
        return 1;
    else return 0;
}

int main()
{
    s=0;
    scanf("%d%d",&n,&m);
    t=2*n+1;
    for(int i=1;i<=n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        ans+=x;
        savex[i]=x;
        savey[i]=y;
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            if(i!=j)
            {
                map[i][j]=MAX; //
            }
            else 
                map[i][j]=0;

    for(int i=0;i<m;i++)
    {
        scanf("%d%d%I64d",&g[i].u,&g[i].v,&g[i].w);
        if(map[g[i].u][g[i].v]>g[i].w)
        {
            map[g[i].u][g[i].v] = map[g[i].v][g[i].u]=g[i].w;
        }
    }
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
            {
                if(map[i][j]>map[i][k]+map[k][j])
                {
                    map[i][j]=map[i][k]+map[k][j];
                }
            }
    __int64 b,d;
    b=0;
    d=50000001000000LL; // 最大的距离，1000000000*200
    if(check(d)!=1)
    {
        printf("-1");
        return 0;
    }
    while(b<d)
    {
        __int64 mid=(b+d)/2;
        if(check(mid)==1)
        {
            d=mid;
        }
        else
        {
            b=mid+1;
        }
    }
    printf("%I64d",b);
    printf("\n");
    return 0;
}