poj 2391 Ombrophobic Bovines(最大流+floyd+二分)

Ombrophobic Bovines

Time Limit: 1000MS
Memory Limit: 65536K

Total Submissions: 14519
Accepted: 3170

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

题意：有F块地，告诉你每块地牛的数量和雨篷能遮蔽的牛的数量，有P条路，告诉你每条路连接的两块地和牛走这条路所需要的时间。

     要你求让所有的牛都能在雨棚下躲雨的最短时间，如果做不到，输出-1

一下解释来自:http://www.2cto.com/kf/201406/312530.html

二分时间，然后把每个田地之间的最短距离用floyd最短路求出来。然后建立一个源点与汇点，将田地拆分成两个点，在距离之内的

进行连边，要单向连边。然后将源点与田地相连，权值为每个田地的牛的数目，再把另一边的田地与汇点相连，权值为每个田地最大

可避雨的牛的数目。拆开的田地之间权值可以为无穷大。

view code#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
typedef long long ll;
const ll INF = 1LL<<60;
const int inf = 1<<30;
const int N = 500;
int n, F, P, pre[N], cur[N];
int s, t, d[N];
ll dis[N][N];

struct om
{
    int num, cap;
}loc[N];

struct edge
{
    int u, v, cap, flow, next;
    edge(int u, int v, int cap, int flow, int next):u(u), v(v), cap(cap), flow(flow), next(next) {}
    edge() {}
}e[N*N*4];
int ecnt;

void floyd()
{
    for(int k=1; k<=n; k++)
        for(int i=1; i<=n; i++) if(dis[i][k]!=INF)
            for(int j=1; j<=n; j++)
                if(dis[k][j]!=INF && dis[i][j]>dis[i][k] + dis[k][j])
                    dis[i][j] = dis[i][k] + dis[k][j];
}

void addedge(int u, int v, int w)
{
    e[ecnt] = edge(u, v, w, 0, pre[u]);
    pre[u] = ecnt++;
    e[ecnt] = edge(v, u, 0, 0, pre[v]);
    pre[v] = ecnt++;
}

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

int DFS(int x, int c)
{
    if(x==t || c==0) return c;
    int flow = 0, f;
    for(int &i=cur[x]; ~i; i=e[i].next)
    {
        int v = e[i].v;
        if(d[x]+1==d[v] && (f=DFS(v,min(c,e[i].cap-e[i].flow)))>0)
        {
            e[i].flow += f;
            e[i^1].flow -=f;
            flow += f;
            c -= f;
            if(c==0) break;
        }
    }
    return flow;
}

ll Maxflow(int s, int t)
{
    int flow = 0;
    while(BFS())
    {
        for(int i=s; i<=t; i++) cur[i] = pre[i];
        flow += DFS(s, inf);
    }
    return flow;
}

bool is_ok(ll m, int sum)
{
    s = 0, t = n*2+1;
    memset(pre, -1, sizeof(pre));
    ecnt = 0;
    for(int i=1; i<=n; i++)
    {
        addedge(s, i, loc[i].num);
        addedge(i+n, t, loc[i].cap);
    }

    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++) if(dis[i][j]<=m)
                addedge(i, j+n, inf);

    return Maxflow(s,t)>=sum;
}

ll solve()
{
    int sumn = 0, sumc = 0;
    for(int i=1; i<=n; i++)
    {
        scanf("%d%d", &loc[i].num, &loc[i].cap);
        sumn += loc[i].num;
        sumc += loc[i].cap;
    }
    int u, v, w;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++) dis[i][j] = INF;
        dis[i][i] = 0;
    }
    for(int i=0; i<P; i++)
    {
        scanf("%d%d%d", &u, &v, &w);
        if(w<dis[u][v]) dis[u][v] = dis[v][u] = w;
    }
    if(sumn > sumc) return -1;
    floyd();

    ll l = 0, r = 0, ans = -1;
    for(int i=1; i<n; i++)
    {
        for(int j=i+1; j<=n; j++)
           if(r<dis[i][j] && dis[i][j]!=INF) r = dis[i][j];
    }
    while(l<=r)
    {
        ll mid = (l+r)>>1;
        if(is_ok(mid, sumn)) ans = mid, r = mid - 1;
        else l = mid + 1;
//        printf("ans = %d\n", ans);
    }
    return ans;
}

int main()
{
//    freopen("in", "r", stdin);
    while(scanf("%d%d", &n, &P)>0) cout<<solve()<<endl;
    return 0;
}

poj 2391 Ombrophobic Bovines(最大流+floyd+二分)