bzoj 3597: [Scoi2014]方伯伯运椰子 0/1分数规划

3597: [Scoi2014]方伯伯运椰子

Time Limit: 30 Sec  Memory Limit: 64 MB
Submit: 144  Solved: 78
[Submit][Status][Discuss]

Description

 .................

Input

 第一行包含二个整数N，M

接下来M行代表M条边，表示这个交通网络

每行六个整数，表示Ui,Vi,Ai,Bi,Ci,Di

接下来一行包含一条边，表示连接起点的边

Output

一个浮点数，保留二位小数。表示答案，数据保证答案大于0

Sample Input

5 10
1 5 13 13 0 412
2 5 30 18 396 148
1 5 33 31 0 39
4 5 22 4 0 786
4 5 13 32 0 561
4 5 3 48 0 460
2 5 32 47 604 258
5 7 44 37 75 164
5 7 34 50 925 441
6 2 26 38 1000 22

Sample Output

103.00

HINT

 1<=N<=5000

0<=M<=3000

1<=Ui,Vi<=N+2

0<=Ai,Bi<=500

0<=Ci<=10000

0<=Di<=1000

 

　　根本搞不懂网上的什么消圈算法，我的思路是这样的：不用想先肯定是枚举答案若ans<delta/tot,那么delta-ans*tot>0，我们先将所有有流的边假设退掉，然后将每条边分为撤销默认操作和新的扩边操作，这样可以直接通过0/1分数规划套费用流解决了，具体怎么加加减减比较烦人，可以直接参考代码。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 5100
#define MAXE MAXV*20
#define MAXV MAXN*2
#define inf 1e100
#define INF 0x3f3f3f3f
#define eps 1e-4
typedef double real;
struct Edge
{
        int np;
        int val;
        real cost;
        Edge *neg,*next;
}E[MAXE],*V[MAXV];
int tope=-1;
int sour=0,sink=1;

void addedge(int x,int y,int v,real c)
{
//        cout<<"Add: "<<x<<" "<<y<<" "<<v<<" "<<c<<endl;
        E[++tope].np=y;
        E[tope].val=v;
        E[tope].cost=c;
        E[tope].next=V[x];
        V[x]=&E[tope];

        E[++tope].np=x;
        E[tope].val=0;
        E[tope].cost=-c;
        E[tope].next=V[y];
        V[y]=&E[tope];

        V[x]->neg=V[y];
        V[y]->neg=V[x];
}
int q[MAXV*20];
int vis[MAXV],vistime;
real dis[MAXV];
Edge *prve[MAXV];
int prv[MAXV];
int n,m;
bool spfa()
{
        int now=sour;
        Edge *ne;
        for (int i=0;i<MAXV;i++)
                dis[i]=-inf;
        dis[now]=0;
        vis[now]=++vistime;
        q[0]=now;
        int head=-1,tail=0;
        while (head<tail)
        {
                now=q[++head];
                vis[now]=0;
                for (ne=V[now];ne;ne=ne->next)
                {
                        if (ne->val && dis[ne->np]<dis[now]+ne->cost)
                        {
                                dis[ne->np]=dis[now]+ne->cost;
                                prv[ne->np]=now;
                                prve[ne->np]=ne;
                                if (vis[ne->np]!=vistime)
                                {
                                        vis[ne->np]=vistime;
                                        q[++tail]=ne->np;
                                }
                        }
                }
        }
        return dis[sink]!=-1e100;
}
pair<int,real> cost_flow()
{
        int x;
        int totf=0;
        real sumc=0;
        int maxf;
        while (spfa())
        {
                maxf=INF;
                for (x=sink;x!=sour;x=prv[x])
                        maxf=min(maxf,prve[x]->val);
                for (x=sink;x!=sour;x=prv[x])
                {
                        prve[x]->val-=maxf;
                        prve[x]->neg->val+=maxf;
                }
                sumc+=maxf*dis[sink];
                totf+=maxf;
        }
        return make_pair(totf,sumc);
}
struct edge
{
        int x,y,vdec,vinc,vcur,vcst;
}e[MAXE];
int main()
{
        //freopen("input.txt","r",stdin);
        int x,y,z;
        scanf("%d%d",&n,&m);
        sour=n+1;
        sink=n+2;
        int mxflow;
        for (int i=0;i<m;i++)
        {
                scanf("%d%d%d%d%d%d",&e[i].x,&e[i].y,&e[i].vdec,&e[i].vinc,&e[i].vcur,&e[i].vcst);
                if (e[i].x==sour)
                        mxflow=e[i].vcur;
        }
        double l,r,mid;
        l=0,r=40000;
        while (l+eps<r)
        {
                memset(V,0,sizeof(V));
                tope=-1;
                mid=(l+r)/2;
                double res=0;
                for (int i=0;i<m;i++)
                {
                        addedge(e[i].x,e[i].y,e[i].vcur,e[i].vdec-e[i].vcst+mid);
                        addedge(e[i].x,e[i].y,mxflow-e[i].vcur,-e[i].vinc-e[i].vcst-mid);
                        res+=e[i].vcur*(-e[i].vdec+e[i].vcst-mid);
                }
                res+=cost_flow().second;
                if (res<=0)
                        r=mid;
                else
                        l=mid;
        }
        printf("%.2lf
",l);
        return 0;
}