POJ 2391 floyd二分+拆点+最大流

Ombrophobic Bovines

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20904		Accepted: 4494

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

Sample Output

解析先求一遍两点之间的最短距离然后二分答案mid，每次二分的时候构建一个网络两点之间的距离<=mid 连一条有向边不过要拆点保证使它是单向的，避免不可达的可达，

跑一边最大流如果等于牛的总数说明mid时间内可以的到达继续二分出最优答案。

我为什么感觉可以费用流解决。。。有时间试一试

#include<iostream>
#include<stdio.h>
#include<vector>
#include<string.h>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn=1e3+20,mod=1e9+7;
const ll inf=1e16;
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("\n");
#define debug(a,b) cout<<a<<" "<<b<<" ";
ll min(ll a,ll b){return a>b?b:a;}
struct edge
{
    int from,to,c,f;
    edge(int u,int v,int c,int f):from(u),to(v),c(c),f(f) {}
};
int n,m,N;
vector<edge> edges;
vector<int> g[maxn];
int d[maxn];//从起点到i的距离
int cur[maxn];//当前弧下标
ll dp[maxn][maxn];
ll a[maxn],b[maxn],sum;
void init(int n)
{
    for(int i=0; i<=N; i++) g[i].clear();
    edges.clear();
}
void addedge(int from,int to,int c) //加边 支持重边
{
    edges.push_back(edge(from,to,c,0));
    edges.push_back(edge(to,from,0,0));
    int siz=edges.size();
    g[from].push_back(siz-2);
    g[to].push_back(siz-1);
}
int bfs(int s,int t) //构造一次层次图
{
    memset(d,-1,sizeof(d));
    queue<int> q;
    q.push(s);
    d[s]=0;
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=0;i<g[x].size();i++)
        {
            edge &e=edges[g[x][i]];
            if(d[e.to]<0&&e.f<e.c) //d[e.to]=-1表示没访问过
            {
                d[e.to]=d[x]+1;
                q.push(e.to);
            }
        }
    }
    return d[t];
}
int dfs(int x,int a,int t) // a表示x点能接收的量
{
    if(x==t||a==0)return a;
    int flow=0,f;//flow总的增量 f一条增广路的增量
    for(int &i=cur[x];i<g[x].size();i++)//cur[i] &引用修改其值 从上次考虑的弧
    {
        edge &e=edges[g[x][i]];
        if(d[x]+1==d[e.to]&&(f=dfs(e.to,min(a,e.c-e.f),t))>0)    //按照层次图增广 满足容量限制
        {
            e.f+=f;
            edges[g[x][i]^1].f-=f;  //修改流量
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    return flow;
}
int maxflow(int s,int t)
{
    int flow=0;
    while(bfs(s,t)!=-1)
    {
        memset(cur,0,sizeof(cur));
        flow+=dfs(s,0x3f3f3f3f,t);
    }
    return flow;
}
void build(ll x)
{
    init(N);
    for(int i=1;i<=n;i++)
    {
        addedge(0,i,a[i]);
        addedge(i+n,N,b[i]);
        addedge(i,i+n,0x3f3f3f3f);
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            if(dp[i][j]<=x)
            {
                addedge(i,j+n,0x3f3f3f3f);
                addedge(j,i+n,0x3f3f3f3f);
            }
        }
    }
}
ll solve()
{
    ll ans=-1;
    ll l=0,r=inf-1;
    while(l<=r)
    {
        ll mid=(l+r)/2;
        build(mid);
        int temp=maxflow(0,N);
        //cout<<mid<<" "<<temp<<endl;
        if(temp>=sum)
        {
            ans=mid;
            r=mid-1;
        }
        else l=mid+1;
    }
    return ans;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        N=n*2+1;sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%ld%ld",&a[i],&b[i]);
            sum+=a[i];
        }
        //====================floyd==========================//
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)  //初始化长度
            {
                if(i==j)
                    dp[i][j]=0;
                else
                    dp[i][j]=inf;
            }
        ll x,y,d;
        for(int i=0;i<m;i++)
        {
            scanf("%lld%lld%lld",&x,&y,&d);
            if(dp[x][y]>d)
                dp[x][y]=dp[y][x]=d;
        }
        for(int k=1; k<=n; k++)
            for(int i=1; i<=n; i++)
                if(dp[i][k]!=inf)
                    for(int j=1; j<=n; j++)
                        dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);
        //=========================================================//
        //init(n);
        cout<<solve()<<endl;
    }
}

POJ 2391 floyd二分+拆点+最大流

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