POJ - 2391 Ombrophobic Bovines (Floyd + 二分 +ISAP)

YOONGI

于 2019-10-07 10:59:53 发布

阅读量201

点赞数

分类专栏：网络流

本文链接：https://blog.youkuaiyun.com/chimchim04/article/details/102291300

版权

网络流专栏收录该内容

6 篇文章

订阅专栏

Ombrophobic Bovines

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

Sample Output

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.

题目链接：http://poj.org/problem?id=2391

题目大意：有F个田地P条路，接下来F行ai，bi表示第i个田地现在有ai头牛，有bi头牛能在这里避雨，接下来是P行ui，vi，wi，表示第i条路在ui和vi之间，牛走过这条路需要花费wi的时间。问所有牛都能避雨要花费的最短时间，如果有牛不能避雨就输出-1

思路：先用Floyd求出每两点间的最小花费时间d[ i ][ j ]，二分花费时间，再建图，拆点，超级源点到点 i 之间建边流量为ai，超级汇点到点 i+F 之间建边流量为bi，如果d[i][j] <= mid, i 到 j+F 建边，流量为inf，然后就是ISAP模板求最大流

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
#define ll long long
const int N=505;
const int M=100005;
const int inf=0x3f3f3f;
const ll INF=0x3f3f3f3f3f3f;
ll mp[N][N];
int g[N],h[N],pre[N];
int first[N],tot,sum,st,ed;
int dis[N],vis[N],a[N],b[N],n,m;
struct node
{
    int v,next;
    int cap,flow;
    node(){};
    node(int tv,int tc,int tnext)
    {
        v=tv,cap=tc,next=tnext,flow=0;
    }
}e[M*2];
void adde(int u,int v,int c)
{
    e[tot]=node(v,c,first[u]);
    first[u]=tot++;
    e[tot]=node(u,0,first[v]);
    first[v]=tot++;
}
void set_h(int t)
{
    memset(h,-1,sizeof(h));
    memset(g,0,sizeof(g));
    queue<int>q;
    h[t]=0;
    q.push(t);
    while(!q.empty())
    {
        int v=q.front();
        q.pop();
        g[h[v]]++;
        for(int i=first[v];~i;i=e[i].next)
        {
            int u=e[i].v;
            if(h[u]==-1)
            {
                h[u]=h[v]+1;
                q.push(u);
            }
        }
    }
}
int ISAP(int s,int t,int n) //ISAP模板
{
    set_h(t);
    int ans=0,u=s,d;
    while(h[s]<n)
    {
        int i=first[u];
        if(u==t) d=inf;
        for(;~i;i=e[i].next)
        {
            int v=e[i].v;
            if(e[i].cap>e[i].flow&&h[u]==h[v]+1)
            {
                u=v;
                pre[v]=i;
                d=min(d,e[i].cap-e[i].flow);
                if(u==t)
                {
                    while(u!=s)
                    {
                        int j=pre[u];
                        e[j].flow+=d;
                        e[j^1].flow-=d;
                        u=e[j^1].v;
                    }
                    ans+=d;
                    d=inf;
                }
                break;
            }
        }
        if(i==-1)
        {
            if(--g[h[u]]==0) break;
            int hmin=n-1;
            for(int j=first[u];~j;j=e[j].next)
                if(e[j].cap>e[j].flow)
                hmin=min(hmin,h[e[j].v]);
            h[u]=hmin+1;
            g[h[u]]++;
            if(u!=s) u=e[pre[u]^1].v;
        }
    }
    return ans;
}
int build(ll w) //建图
{
    tot=0;
    memset(first,-1,sizeof(first));
    st=0,ed=2*n+1; //超级源点汇点
    for(int i=1; i<=n; i++)
    {
        adde(st,i,a[i]);
        adde(i+n,ed,b[i]);
        for(int j=1; j<=n; j++)
            if(mp[i][j]<=w)
                adde(i,j+n,inf);
    }
    return ISAP(st,ed,2*n+2);
}
void solve()
{
    ll maxx=0;
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
            if(mp[i][j]<INF&&mp[i][j]>maxx)
                maxx=mp[i][j];
    if(build(maxx)<sum) printf("-1\n");
    else
    {
        ll l=0,r=maxx,ans;
        while(l<=r)
        {
            ll mid=(l+r)>>1;
            if(build(mid)>=sum)
            {
                ans=mid;
                r=mid-1;
            }
            else l=mid+1;
        }
        printf("%lld\n",ans);
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    sum=0;
    int u,v;
    ll w;
    for(int i=1; i<=n; i++)
    {
        scanf("%d%d",&a[i],&b[i]);
        sum+=a[i];
    }
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
            if(i==j) mp[i][j]=0;
            else mp[i][j]=INF;
    }
    for(int i=0; i<m; i++)
    {
        scanf("%d%d%lld",&u,&v,&w);
        if(w<mp[u][v]) mp[u][v]=mp[v][u]=w;
    }
    for(int k=1; k<=n; k++) //Floyd
        for(int i=1; i<=n; i++)
            if(mp[i][k]<INF)
                for(int j=1; j<=n; j++)
                    mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
    solve();
    return 0;
}