POJ2391 Ombrophobic Bovines（网络流）（拆点）

时间：2016-09-02 21:39:27 阅读：130 评论：0 收藏：0 [点我收藏+]

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Ombrophobic Bovines

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18205		Accepted: 3960

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

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.

【分析】这个题跟POJ2112很像，不过2112求的是最小的单条路，而这个题求的是最小的路径长度，所以要拆点，剩下的就是网络流了。一开始一直WA，后来把cost的初始化和Floyd改了一下就过了，感觉两种写法没什么区别啊，求大神指教。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=505;
const int M=300005;
int power(int a,int b,int c){int ans=1;while(b){if(b%2==1){ans=(ans*a)%c;b--;}b/=2;a=a*a%c;}return ans;}
struct man
{
    int c,f;
}w[N][N];
int dis[N],n,m;
int t,cnt,maxn=0,ans;
ll cost[N][N];
int c[N],f[N];
bool bfs()
{
    queue<int>q;
    memset(dis,0,sizeof(dis));
    q.push(0);
    dis[0]=1;
    while(!q.empty() && !dis[t]){
        int v=q.front();q.pop();
        for(int i=1;i<=t;i++){
                //if(i==t)printf("w[i][t].c=%d\n",w[i][t].c);
            if(!dis[i]&&w[v][i].c>w[v][i].f){
                q.push(i);
                dis[i]=dis[v]+1;
            }
        }
    }
    return dis[t]!=0;
}
int dfs(int cur,int cp)
{
    if(cur==t||cp==0)return cp;
    int tmp=cp,tt;
    for(int i=1;i<=t;i++){
        if(dis[i]==dis[cur]+1 &&w[cur][i].c>w[cur][i].f){
            tt=dfs(i,min(w[cur][i].c-w[cur][i].f,tmp));
            w[cur][i].f+=tt;
            w[i][cur].f-=tt;
            tmp-=tt;
        }
    }
    return cp-tmp;
}
void dinic()
{
    ans=0;
    while(bfs())ans+=dfs(0,inf);
}
void Floyd()
{
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(i!=j)cost[i][j]=min(cost[i][j],cost[i][k]+cost[k][j]);
                else cost[i][j]=0;
            }
        }
    }
}
void Build(ll x)
{
    memset(w,0,sizeof(w));
    for(int i=1;i<=n;i++)w[0][i].c=c[i];
    for(int i=n+1;i<=2*n;i++)w[i][t].c=f[i-n];
    for(int i=1;i<=n;i++)for(int j=n+1;j<t;j++)if(cost[i][j-n]<=x)w[i][j].c=inf;
}
int main(){
    cin>>n>>m;
    memset(cost,inf,sizeof(cost));
    ll l=0,r=1;
    t=n*2+1;
    for(int i=1;i<=n;i++){cin>>c[i]>>f[i];maxn+=c[i];}
    int a,b;ll val;
    while(m--){
        cin>>a>>b>>val;
        r+=val;
        cost[a][b]=cost[b][a]=min(cost[a][b],val);
    }
    Floyd();
    bool flag=false;
    while(l<r){
        ll mid=(l+r)/2;
        Build(mid);
        dinic();
        if(ans>=maxn)r=mid,flag=true;
        else l=mid+1;
    }
    if(flag) cout<<r<<endl;
    else puts("-1");
    return 0;
}

AC代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=505;
const int M=300005;
int power(int a,int b,int c){int ans=1;while(b){if(b%2==1){ans=(ans*a)%c;b--;}b/=2;a=a*a%c;}return ans;}
struct man
{
    int c,f;
}w[N][N];
int dis[N],n,m;
int t,cnt,maxn=0,ans;
ll cost[N][N];
int c[N],f[N];
bool bfs()
{
    queue<int>q;
    memset(dis,0,sizeof(dis));
    q.push(0);
    dis[0]=1;
    while(!q.empty() && !dis[t]){
        int v=q.front();q.pop();
        for(int i=1;i<=t;i++){
                //if(i==t)printf("w[i][t].c=%d\n",w[i][t].c);
            if(!dis[i]&&w[v][i].c>w[v][i].f){
                q.push(i);
                dis[i]=dis[v]+1;
            }
        }
    }
    return dis[t]!=0;
}
int dfs(int cur,int cp)
{
    if(cur==t||cp==0)return cp;
    int tmp=cp,tt;
    for(int i=1;i<=t;i++){
        if(dis[i]==dis[cur]+1 &&w[cur][i].c>w[cur][i].f){
            tt=dfs(i,min(w[cur][i].c-w[cur][i].f,tmp));
            w[cur][i].f+=tt;
            w[i][cur].f-=tt;
            tmp-=tt;
        }
    }
    return cp-tmp;
}
void dinic()
{
    ans=0;
    while(bfs())ans+=dfs(0,inf);
}
void Floyd()
{
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            if(cost[i][k]!=inf){
                for(int j=1;j<=n;j++){
                    cost[i][j]=min(cost[i][j],cost[i][k]+cost[k][j]);
                }
            }
        }
    }
}
void Build(ll x)
{
    memset(w,0,sizeof(w));
    for(int i=1;i<=n;i++)w[0][i].c=c[i];
    for(int i=n+1;i<=2*n;i++)w[i][t].c=f[i-n];
    for(int i=1;i<=n;i++)for(int j=n+1;j<t;j++)if(cost[i][j-n]<=x)w[i][j].c=inf;
}
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(i==j)cost[i][j]=0;
            else cost[i][j]=inf;
        }
    }
    ll l=0,r=1;
    t=n*2+1;
    for(int i=1;i<=n;i++){cin>>c[i]>>f[i];maxn+=c[i];}
    int a,b;ll val;
    while(m--){
        cin>>a>>b>>val;
        r+=val;
        cost[a][b]=cost[b][a]=min(cost[a][b],val);
    }
    Floyd();
    bool flag=false;
    while(l<r){
        ll mid=(l+r)/2;
        Build(mid);
        dinic();
        if(ans>=maxn)r=mid,flag=true;
        else l=mid+1;
    }
    if(flag) cout<<r<<endl;
    else puts("-1");
    return 0;
}

View Code

上面的是WA代码，不知道为什么错了。

POJ2391 Ombrophobic Bovines（网络流）（拆点）

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原文地址：http://www.cnblogs.com/jianrenfang/p/5835562.html

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