POJ2391 Ombrophobic Bovines

Ombrophobic Bovines

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19359		Accepted: 4186

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

———————————————————————————————

题目的意思是给出n个牛棚的牛数量和容量及各个牛棚间距离，求让所有牛找找碰最短距离

思路：先floyd求各个点最短距离，再二分最短距离用网络流验证，见图示要把一个牛棚拆成2个点。

别忘了long long和不行输出-1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
#define MAXN 500

struct node
{
    int u, v, next, cap;
} edge[MAXN*MAXN];
int nt[MAXN], s[MAXN], d[MAXN], visit[MAXN];
int cnt;
int n,m,k;
LL mp[MAXN][MAXN];
int a[MAXN],b[MAXN];

void init()
{
    cnt = 0;
    memset(s, -1, sizeof(s));
}

void add(int u, int v, int c)
{
    edge[cnt].u = u;
    edge[cnt].v = v;
    edge[cnt].cap = c;
    edge[cnt].next = s[u];
    s[u] = cnt++;
    edge[cnt].u = v;
    edge[cnt].v = u;
    edge[cnt].cap = 0;
    edge[cnt].next = s[v];
    s[v] = cnt++;
}

bool BFS(int ss, int ee)
{
    memset(d, 0, sizeof d);
    d[ss] = 1;
    queue<int>q;
    q.push(ss);
    while (!q.empty())
    {
        int pre = q.front();
        q.pop();
        for (int i = s[pre]; ~i; i = edge[i].next)
        {
            int v = edge[i].v;
            if (edge[i].cap > 0 && !d[v])
            {
                d[v] = d[pre] + 1;
                q.push(v);
            }
        }
    }
    return d[ee];
}

int DFS(int x, int exp, int ee)
{
    if (x == ee||!exp) return exp;
    int temp,flow=0;
    for (int i = nt[x]; ~i ; i = edge[i].next, nt[x] = i)
    {
        int v = edge[i].v;
        if (d[v] == d[x] + 1&&(temp = (DFS(v, min(exp, edge[i].cap), ee))) > 0)
        {
            edge[i].cap -= temp;
            edge[i ^ 1].cap += temp;
            flow += temp;
            exp -= temp;
            if (!exp) break;
        }
    }
    if (!flow) d[x] = 0;
    return flow;
}

int Dinic_flow(LL mid)
{
    init();

    for(int i=1; i<=n; i++)
        add(0,i,a[i]),add(i+n,2*n+1,b[i]);
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
            if(mp[i][j]<=mid)
                add(i,j+n,INF);
    int ss=0,ee=2*n+1;
    int ans = 0;
    while (BFS(ss, ee))
    {
        for (int i = 0; i <=ee; i++) nt[i] = s[i];
        ans+= DFS(ss, INF, ee);
    }
    return ans;;
}

void floyd()
{
    for(int k = 1; k <= n; ++k)
    {
        for(int i = 1; i <= n; ++i)
        {
            for(int j = 1; j <= n; ++j)
            {
                mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);
            }
        }
    }
}

int main()
{
    int u,v;
    LL c;
    while(~scanf("%d%d",&n,&m))
    {

        for(int i=0; i<=n; i++)
            for(int j=0; j<=n; j++)
            {
                if(i!=j)
                    mp[i][j]=1e14;
                else
                    mp[i][j]=0;
            }
   int sum=0;
        for(int i=1; i<=n; i++)
            scanf("%d%d",&a[i],&b[i]),sum+=a[i];
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%lld",&u,&v,&c);
            mp[u][v]=min(c,mp[u][v]);
            mp[v][u]=mp[u][v];
        }
        floyd();
        LL l=0,r=1e13;
        LL ans=-1;
        while(l<=r)
        {
            LL mid=(l+r)/2;
            if(Dinic_flow(mid)==sum) ans=mid,r=mid-1;
            else l=mid+1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}