POJ 2391 Ombrophobic Bovines

Ombrophobic Bovines

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18623		Accepted: 4057

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：

f个草坪，每个草坪初始有a[i]头牛，最多可以容纳b[i]头牛，无向图，问最少需要多少时间可以使得每头牛都有归宿...

分析：

最大流的基础题目...但是我貌似脑残了...TAT...

先Floyd处理出每两个点之间的最短路，二分答案，然后建图...

我们第一想法一定是拆点，把每个点拆成一个出点一个入点，S向入点连一条容量为a[i]的边，出点向T连一条容量为b[i]的边，如果两个点之间最短路小于枚举的ans就连边...

但是这肯定是错误的...(随便一个数据就可以卡...)

正确的建图方法是S向出点连边，入点向T连边，出点向入点连边...这样一头牛从A转移到B之后就不可能再转移到其他点了...

zz的我把lr定义成了long long但是忘记改mid...TAT...

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
//by NeighThorn
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
using namespace std;

const int maxn=200+5,maxm=100000+5;

int n,m,S,T,cnt,sum,a[maxn],b[maxn],hd[maxn*2],fl[maxm],to[maxm],nxt[maxm],pos[maxn*2];
long long dis[maxn][maxn],Max;

inline void add(int s,int x,int y){
    fl[cnt]=s;to[cnt]=y;nxt[cnt]=hd[x];hd[x]=cnt++;
    fl[cnt]=0;to[cnt]=x;nxt[cnt]=hd[y];hd[y]=cnt++;
}

inline bool bfs(void){
    memset(pos,-1,sizeof(pos));
    int head=0,tail=0,q[maxn*2];
    q[0]=S,pos[S]=0;
    while(head<=tail){
        int top=q[head++];
        for(int i=hd[top];i!=-1;i=nxt[i])
            if(pos[to[i]]==-1&&fl[i])
                pos[to[i]]=pos[top]+1,q[++tail]=to[i];
    }
    return pos[T]!=-1;
}

inline int find(int v,int f){
    if(v==T)
        return f;
    int res=0,t;
    for(int i=hd[v];i!=-1&&f>res;i=nxt[i])
        if(pos[to[i]]==pos[v]+1&&fl[i])
            t=find(to[i],min(fl[i],f-res)),fl[i]-=t,fl[i^1]+=t,res+=t;
    if(!res)
        pos[v]=-1;
    return res;
}

inline int dinic(void){
    int res=0,t;
    while(bfs())
        while(t=find(S,inf))
            res+=t;
    return res;
}

inline int check(long long mid){
    cnt=0;memset(hd,-1,sizeof(hd));
    for(int i=1;i<=n;i++)
        add(a[i],S,i+n),add(b[i],i,T),add(inf,i+n,i);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            if(dis[i][j]<=mid)
                add(inf,i+n,j);
    return dinic();
}

signed main(void){
//    freopen("in.txt","r",stdin);
    Max=0,sum=cnt=0;
    scanf("%d%d",&n,&m);
    S=0,T=n*2+1;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            dis[i][j]=INF;
    for(int i=1;i<=n;i++)
        scanf("%d%d",&a[i],&b[i]),sum+=a[i];
    for(int i=1,s,x,y;i<=m;i++)
        scanf("%d%d%d",&x,&y,&s),dis[x][y]=dis[y][x]=min(dis[x][y],(long long)s);
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
    long long l=0,r=INF-1,ans=-1;
    while(l<=r){
        long long mid=(l+r)>>1;
        if(check(mid)==sum)
            ans=mid,r=mid-1;
        else
            l=mid+1;
    }
    printf("%lld
",ans);
    return 0;  
}//Cap ou pas cap. Cap.

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By NeighThorn