Poj 2391 ombrophobic bovines (maximum stream + Floyd + binary)

Ombrophobic bovines

Time limit:1000 ms
Memory limit:65536 K

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 47 20 42 61 2 403 2 702 3 901 3 120

Sample output

110

There are f places, which tell you the number of cattle each and the number of cattle that can be covered by the awning, there are P ways, it tells you the time required to connect two locations and take the same route with the ox.

If you want to keep all the cows in the shortest time in the canopy, output-1

From: http://www.2cto.com/kf/201406/312530.html

The second time, and then the shortest distance between each field is obtained using the Floyd shortest path. Create a source point and sink point, split the field into two points, within the distance

One-way EDGE connection is required. Then the source point is connected to the field, the weight is the number of cattle in each field, and the other side of the field is connected to the sink, the weight is the maximum

Number of cattle that can withstand the rain. The split field can have an infinite weight.

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;}