BZOJ1738: [Usaco2005 mar]Ombrophobic Bovines 发抖的牛

n<=200个点m<=1500条无向带权边的图，每个点有人和容量，人可以移动，代价为所有人走过的边的权和，求使所有点人不超过容量的最小代价。

方法一：费用流。

错误！答案与边权不成比例。

方法二：二分一个答案，然后根据floyd求出的最短路看每个点在二分的答案下能去到哪些点，跑最大流检查是否合法。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
//#include<iostream>
using namespace std;

int n,m;
#define maxn 411
#define maxm 100011
#define LL long long
int a[maxn],b[maxn],sum=0;
LL mp[maxn][maxn];
const LL inf=1e15;
struct Edge{int from,to,next,cap,flow;};
struct Network
{
    Edge edge[maxm];int n,s,t,le;
    int first[maxn],dis[maxn],cur[maxn];
    void clear(int n)
    {
        memset(first,0,sizeof(first));
        le=2;this->n=n;
    }
    void add_edge(int x,int y,int v)
    {
        Edge &e=edge[le];
        e.to=y;e.from=x;
        e.cap=v;e.flow=0;
        e.next=first[x];
        first[x]=le++;
    }
    void insert(int x,int y,int v)
    {
        add_edge(x,y,v);
        add_edge(y,x,0);
    }
    int que[maxn],head,tail;
    bool bfs()
    {
        memset(dis,0,sizeof(dis));
        dis[s]=1;
        que[head=(tail=1)-1]=s;
        while (head!=tail)
        {
            const int x=que[head++];
            for (int i=first[x];i;i=edge[i].next)
            {
                Edge &e=edge[i];
                if (e.cap>e.flow && !dis[e.to])
                {
                    dis[e.to]=dis[x]+1;
                    que[tail++]=e.to;
                }
            }
        }
        return dis[t];
    }
    int dfs(int x,int a)
    {
        if (x==t || !a) return a;
        int flow=0,f;
        for (int &i=cur[x];i;i=edge[i].next)
        {
            Edge &e=edge[i];
            if (dis[e.to]==dis[x]+1 && (f=dfs(e.to,min(a,e.cap-e.flow)))>0)
            {
                flow+=f;
                a-=f;
                e.flow+=f;
                edge[i^1].flow-=f;
                if (!a) break;
            }
        }
        return flow;
    }
    int Dinic(int s,int t)
    {
        this->s=s;this->t=t;
        int ans=0;
        while (bfs())
        {
            for (int i=1;i<=n;i++) cur[i]=first[i];
            ans+=dfs(s,0x3f3f3f3f);
        }
        return ans;
    }
}g;
bool check(LL x)
{
    g.clear(n*2+2);
    int s=g.n-1,t=g.n;
    for (int i=1;i<=n;i++)
    {
        g.insert(s,i,a[i]);
        g.insert(i+n,t,b[i]);
    }
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
            if (mp[i][j]<=x) g.insert(i,j+n,0x3f3f3f3f);
    return g.Dinic(s,t)==sum;
}
int x,y,v;
int main()
{
    scanf("%d%d",&n,&m);
    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++)
            mp[i][j]=inf;
    for (int i=1;i<=n;i++) mp[i][i]=0;
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&v);
        mp[x][y]=min(mp[x][y],(LL)v);
        mp[y][x]=min(mp[y][x],(LL)v);
    }
    for (int k=1;k<=n;k++)
        for (int i=1;i<=n;i++)
            for (int j=1;j<=n;j++)
                if (mp[i][j]>mp[i][k]+mp[k][j])
                    mp[i][j]=mp[i][k]+mp[k][j];
    LL L=0,R=1e15;
    while (L<R)
    {
        const LL mid=(L+R)>>1;
        if (check(mid)) R=mid;
        else L=mid+1;
    }
    printf(R==1e15?"-1\n":"%lld\n",L);
    return 0;
}