POJ3621Sightseeing Cows[01分数规划 spfa(dfs)负环 ]

Sightseeing Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9703		Accepted: 3299

Description

Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time.

Fortunately, they have a detailed city map showing the L (2 ≤ L ≤ 1000) major landmarks (conveniently numbered 1.. L) and the P (2 ≤ P ≤ 5000) unidirectional cow paths that join them. Farmer John will drive the cows to a starting landmark of their choice, from which they will walk along the cow paths to a series of other landmarks, ending back at their starting landmark where Farmer John will pick them up and take them back to the farm. Because space in the city is at a premium, the cow paths are very narrow and so travel along each cow path is only allowed in one fixed direction.

While the cows may spend as much time as they like in the city, they do tend to get bored easily. Visiting each new landmark is fun, but walking between them takes time. The cows know the exact fun values F_i (1 ≤ F_i ≤ 1000) for each landmark i.

The cows also know about the cowpaths. Cowpath i connects landmark L_1i to L_2i (in the direction L_1i -> L_2i ) and requires time T_i (1 ≤ T_i ≤ 1000) to traverse.

In order to have the best possible day off, the cows want to maximize the average fun value per unit time of their trip. Of course, the landmarks are only fun the first time they are visited; the cows may pass through the landmark more than once, but they do not perceive its fun value again. Furthermore, Farmer John is making the cows visit at least two landmarks, so that they get some exercise during their day off.

Help the cows find the maximum fun value per unit time that they can achieve.

Input

* Line 1: Two space-separated integers: L and P
* Lines 2..L+1: Line i+1 contains a single one integer: F_i
* Lines L+2..L+P+1: Line L+i+1 describes cow path i with three space-separated integers: L_1i , L_2i , and T_i

Output

* Line 1: A single number given to two decimal places (do not perform explicit rounding), the maximum possible average fun per unit time, or 0 if the cows cannot plan any trip at all in accordance with the above rules.

Sample Input

5 7
30
10
10
5
10
1 2 3
2 3 2
3 4 5
3 5 2
4 5 5
5 1 3
5 2 2

Sample Output

6.00

Source

USACO 2007 December Gold

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转载一段题解：

http://www.cnblogs.com/wally/p/3228171.html
题目的意思是：求一个环的{点权和}除以{边权和}，使得那个环在所有环中{点权和}除以{边权和}最大。
令在一个环里，点权为v[i]，对应的边权为e[i]， 
即要求：∑（i=1，n）v[i]/∑（i=1，n)e[i]最大的环（n为环的点数）， 
设题目答案为ans， 
即对于所有的环都有 ∑（i=1，n）（v[i]）/∑（i=1，n)（e[i]）<=ans 
变形得ans* ∑（i=1，n）（e[i]）>=∑（i=1，n）（v[i]） 
再得 ∑（i=1，n）（ans*e[i]-v[i]） >=0 
稍分析一下，就有： 
当k<ans时，就存在至少一个环∑（i=1，n）（k*e[i]-v[i]）<0，即有负权回路(边权为k*e[i]-v[i])； 
当k>=ans时，就对于所有的环∑（i=1，n）（k*e[i]-v[i]）>=0，即没有负权回路。 
然后我们就可以使新的边权为k*e[i]-v[i]，用spfa来判断付权回路,二分ans。

再一段

http://www.cnblogs.com/proverbs/archive/2013/01/09/2853741.html

01分数规划，简单构造，将点权转移到边权上~因为一个环上的点和边的数量是相等的~

设i,j之间初始边权为w[i][j]，修改后的边权为g[i][j]，则g[i][j]=w[i][j]*mid+val[i]

spfa判负环即可~

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列式转换一下，所有的都满足

 ∑（i=1，n）（ans*e[i]-v[i]） >=0
如果有的不满足，就更改ans使他满足，用实数二分的方法

用spfa判负环，有负环就说明可以选出一个n的集合使上述条件不满足
因为不需要最短路所以把初始化所有点加进去d=0
还有一种很神的dfs写法，负环超级快；从每个点dfs-spfa看看有没有回到自己

bfs：

//
//  main.cpp
//  poj3621
//
//  Created by Candy on 9/13/16.
//  Copyright ? 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
const int N=1005,M=5005;

int n,m,a,b;
double l,r,mid,f[N],c;
struct edge{
    int v,ne;
    double w;
}e[M];
int h[N],cnt=0;
inline void ins(int u,int v,double w){
    cnt++;
    e[cnt].v=v;e[cnt].w=w;e[cnt].ne=h[u];h[u]=cnt;
}

int inq[N],num[N];
double d[N];
bool spfanc(double p){
    queue<int> q;
    memset(num,0,sizeof(num));
    for(int i=1;i<=n;i++) {d[i]=0;inq[i]=1;q.push(i);num[i]++;}
    
    while(!q.empty()){
        int u=q.front();q.pop();inq[u]=0;
        for(int i=h[u];i;i=e[i].ne){
            int v=e[i].v; double w=e[i].w*p-f[u];
            if(d[v]>d[u]+w){
                d[v]=d[u]+w;
                if(!inq[v]){q.push(v);inq[v]=1;if(++num[v]>n) return true;}
            }
        }
    }
    return false;
}
int main(int argc, const char * argv[]) {
    while(cin>>n>>m){
        cnt=0;
        memset(h,0,sizeof(h));
        for(int i=1;i<=n;i++) scanf("%lf",&f[i]);
    for(int i=1;i<=m;i++){scanf("%d%d%lf",&a,&b,&c);ins(a,b,c);}
    
    l=0,r=2000;
    while(r-l>1e-4){
        mid=(l+r)/2;//printf("%.2f\n",mid);
        if(spfanc(mid)) l=mid;
        else r=mid;
    }
    printf("%.2f\n",mid);
    }
    return 0;
}

 dfs：

//
//  main.cpp
//  poj3621
//
//  Created by Candy on 9/13/16.
//  Copyright ? 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
const int N=1005,M=5005;

int n,m,a,b;
double l,r,mid,f[N],c;
struct edge{
    int v,ne;
    double w;
}e[M];
int h[N],cnt=0;
inline void ins(int u,int v,double w){
    cnt++;
    e[cnt].v=v;e[cnt].w=w;e[cnt].ne=h[u];h[u]=cnt;
}

int vis[N],st;
double d[N];
inline bool dfs(int u,double p){
    vis[u]=st;
    for(int i=h[u];i;i=e[i].ne){
        int v=e[i].v; double w=e[i].w*p-f[u];
        if(d[v]>d[u]+w){
            d[v]=d[u]+w;
            if(vis[v]==vis[u]) return true;
            else if(dfs(v,p)) return true;
        }
    }
    vis[u]=0;
    return false;
}
bool nc(double p){
    memset(vis,0,sizeof(vis));
    //memset(d,0,sizeof(d));
    for(st=1;st<=n;st++)
        if(dfs(st,p)) return true;
    
    return false;
}
int main(int argc, const char * argv[]) {
    while(cin>>n>>m){
        cnt=0;
        memset(h,0,sizeof(h));
        for(int i=1;i<=n;i++) scanf("%lf",&f[i]);
        for(int i=1;i<=m;i++){scanf("%d%d%lf",&a,&b,&c);ins(a,b,c);}
        
        l=0,r=1000;
        while(r-l>1e-4){
            mid=(l+r)/2;//printf("%.2f\n",mid);
            if(nc(mid)) l=mid;
            else r=mid;
        }
        printf("%.2f\n",mid);
    }
    return 0;
}