POJ 2455 Secret Milking Machine（搜索-二分,网络流-最大流）

深度探讨网络流算法在远距离路径规划中的应用

最新推荐文章于 2018-11-24 23:24:00 发布

转载最新推荐文章于 2018-11-24 23:24:00 发布 · 61 阅读

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原文链接：http://www.cnblogs.com/toyking/p/4951010.html

本文详细介绍了如何运用网络流算法解决从起点到终点的多路径规划问题，确保路径中每段路程长度最优化，同时避免重复使用同一路段。通过实例分析，展示了算法在复杂农场环境下的实际应用，旨在最小化最长路径长度，提高效率并降低被发现的风险。

Secret Milking Machine

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9658		Accepted: 2859

Description

Farmer John is constructing a new milking machine and wishes to keep it secret as long as possible. He has hidden in it deep within his farm and needs to be able to get to the machine without being detected. He must make a total of T (1 <= T <= 200) trips to the machine during its construction. He has a secret tunnel that he uses only for the return trips.

The farm comprises N (2 <= N <= 200) landmarks (numbered 1..N) connected by P (1 <= P <= 40,000) bidirectional trails (numbered 1..P) and with a positive length that does not exceed 1,000,000. Multiple trails might join a pair of landmarks.

To minimize his chances of detection, FJ knows he cannot use any trail on the farm more than once and that he should try to use the shortest trails.

Help FJ get from the barn (landmark 1) to the secret milking machine (landmark N) a total of T times. Find the minimum possible length of the longest single trail that he will have to use, subject to the constraint that he use no trail more than once. (Note well: The goal is to minimize the length of the longest trail, not the sum of the trail lengths.)

It is guaranteed that FJ can make all T trips without reusing a trail.

Input

* Line 1: Three space-separated integers: N, P, and T

* Lines 2..P+1: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, indicating that a trail connects landmark A_i to landmark B_i with length L_i.

Output

* Line 1: A single integer that is the minimum possible length of the longest segment of Farmer John's route.

Sample Input

Sample Output

Hint

Farmer John can travel trails 1 - 2 - 3 - 7 and 1 - 6 - 7. None of the trails travelled exceeds 5 units in length. It is impossible for Farmer John to travel from 1 to 7 twice without using at least one trail of length 5.

Huge input data,scanf is recommended.

Source

USACO 2005 February Gold

题目大意：

FJ有N块地，这些地之间有P条双向路，每条路的都有固定的长度l。现在要你找出从第1块地到第n块地的T条不同路径，每条路径上的路不能与先前的路径重复，问这些路径中的最长路的最小是多少。

解题思路：

二分+网络流。

解题代码：

代码一：DINIC算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;

const int INF=(1<<30);
const int maxn=210,maxm=201000;

struct edge{
    int u,v,f,next;
    edge(int u0=0,int v0=0,int f0=0){
        u=u0;v=v0;f=f0;
    }
}e[maxm];

int src,sink,cnt,head[maxn];

void adde(int u,int v,int f){
    e[cnt].u=u,e[cnt].v=v,e[cnt].f=f,e[cnt].next=head[u],head[u]=cnt++;
    e[cnt].u=v,e[cnt].v=u,e[cnt].f=0,e[cnt].next=head[v],head[v]=cnt++;
}

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

queue <int> q;
bool visited[maxn];
int dist[maxn];

void bfs(){
    memset(dist,0,sizeof(dist));
    while(!q.empty()) q.pop();
    visited[src]=true;
    q.push(src);
    while(!q.empty()){
        int s=q.front();
        q.pop();
        for(int i=head[s];i!=-1;i=e[i].next){
            int d=e[i].v;
            if(e[i].f>0 && !visited[d]){
                q.push(d);
                dist[d]=dist[s]+1;
                visited[d]=true;
            }
        }
    }
}

int dfs(int u,int delta){
    if(u==sink) return delta;
    else{
        int ret=0;
        for(int i=head[u];delta && i!=-1;i=e[i].next){
            if(e[i].f>0 && dist[e[i].v]==dist[u]+1){
                int d=dfs(e[i].v,min(e[i].f,delta));
                e[i].f-=d;
                e[i^1].f+=d;
                delta-=d;
                ret+=d;
            }
        }
        if(!ret) dist[u]=-2;
        return ret;
    }
}

int maxflow(){
    int ret=0;
    while(true){
        memset(visited,false,sizeof(visited));
        bfs();
        if(!visited[sink]) return ret;
        ret+=dfs(src,INF);
    }
    return ret;
}

int n,m,num,maxr,minr;
vector <edge> vec;

void input(){
    maxr=0;
    minr=INF;
    src=1,sink=n;
    vec.clear();
    int u,v,w;
    for(int i=0;i<m;i++){
        scanf("%d%d%d",&u,&v,&w);
        vec.push_back(edge(u,v,w));
        vec.push_back(edge(v,u,w));
        if(w>maxr) maxr=w;
        if(w<minr) minr=w;
    }
}

void build(int dis0){
    init();
    for(int i=0;i<vec.size();i++){
        if(vec[i].f<=dis0){
            adde(vec[i].u,vec[i].v,1);
        }
    }
}

void solve(){
    int l=minr,r=maxr;
    while(l<r){
        int mid=(l+r)/2;
        build(mid);
        if(maxflow()>=num) r=mid;
        else l=mid+1;
    }
    printf("%d\n",r);
}

int main(){
    while(scanf("%d%d%d",&n,&m,&num)!=EOF){
        input();
        solve();
    }
    return 0;
}

代码二：sap算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

#define INF 2000000000
#define N 100010
typedef long long ll;

const int maxn=210;
struct edge{
    int u,v,next,cap;
    edge(int u0=0,int v0=0,int f0=0){
        u=u0;v=v0;cap=f0;
    }
}e[210000];

int n,head[N],tol,top,st[N];
int src,des,dep[N],gap[N];

void adde(int u,int v,int c){
    e[tol].u=u,e[tol].v=v,e[tol].next=head[u],e[tol].cap=c,head[u]=tol++;
    e[tol].u=v,e[tol].v=u,e[tol].next=head[v],e[tol].cap=0,head[v]=tol++;
}

void bfs(){//对于反边计算层次
    for(int i=0;i<N;i++) dep[i]=N-1;
    memset(gap,0,sizeof gap);
    gap[0]=1,dep[des]=0;
    int q[N],l=0,r=0,u,v;
    q[r++]=des;
    while(l!=r){
        u=q[l++];
        l=l%N;
        for(int i=head[u];i!=-1;i=e[i].next){
            v=e[i].v;
            if(e[i].cap!=0||dep[v]!=N-1) continue;
            q[r++]=v;
            r=r%N;
            ++gap[dep[v]=dep[u]+1];
        }
    }
}

int sap(){
    bfs();
    int u=src,s[N],top=0,res=0,ii;
    int cur[N];
    memcpy(cur,head,sizeof head);
    while(dep[src]<n){
        if(u==des){//求得一条增广路
           int minf=INF,pos=n;
           for(int i=0;i<top;i++){
              if(minf>e[s[i]].cap){
                  minf=e[s[i]].cap;
                  pos=i;
              }
           }
           for(int i=0;i<top;i++){
              e[s[i]].cap-=minf;
              e[s[i]^1].cap+=minf;
           }
           top=pos;
           res+=minf;
           u=e[s[top]].u;//优化1
        }
        if(dep[u]!=0&&gap[dep[u]-1]==0) break;//出现断层
        ii=-1;
        for(int i=cur[u];i!=-1;i=e[i].next){
             if(dep[e[i].v]==N-1) continue;
             if(e[i].cap!=0&&dep[u]==dep[e[i].v]+1){ii=i;break;}
        }
        if(ii!=-1){//有允许弧
            cur[u]=ii;
            s[top++]=ii;
            u=e[ii].v;
        }else{//不断回退找增光路
            int mind=n;
            for(int i=head[u];i!=-1;i=e[i].next){
                if(e[i].cap==0) continue;
                if(dep[e[i].v]<mind) mind=dep[e[i].v],cur[u]=i;
            }
            --gap[dep[u]];
            ++gap[dep[u]=mind+1];//优化2
            if(u!=src) u=e[s[--top]].u;
        }
    }
    return res;
}

int m,num,maxr,minr;
vector <edge> vec;

void input(){
    maxr=0;
    minr=INF;
    vec.clear();
    int u,v,w;
    for(int i=0;i<m;i++){
        scanf("%d%d%d",&u,&v,&w);
        vec.push_back(edge(u,v,w));
        vec.push_back(edge(v,u,w));
        if(w>maxr) maxr=w;
        if(w<minr) minr=w;
    }
}

void build(int dis0){
    tol=0;
    memset(head,-1,sizeof head);
    src=1,des=n,n;
    int vsize=vec.size();
    for(int i=0;i<vsize;i++){
        if(vec[i].cap<=dis0) adde(vec[i].u,vec[i].v,1);
    }
}

void solve(){
    int l=minr,r=maxr;
    while(l<r){
        int mid=(l+r)/2;
        build(mid);
        if(sap()>=num) r=mid;
        else l=mid+1;
    }
    printf("%d\n",r);
}

int main(){
    while(scanf("%d%d%d",&n,&m,&num)!=EOF){
        input();
        solve();
    }
    return 0;
}

转载于:https://www.cnblogs.com/toyking/p/4951010.html