ZOJ2314 Reactor Cooling (有上下界的网络流)

来源：互联网发布：字幕组怎么赚钱知乎编辑：程序博客网时间：2024/05/09 10:27

sort貌似不能给2维数组排序呢？改成qsort就好了，这里的构图方法和周源的论文里的类似

题目是求一个有上下界网络中的无源汇的可行流。

构造伴随网络方法如下：

新增两个顶点VS VT 对原网络每个顶点算其D(u)的值，其中D(u)为顶点u发出的所有弧的流量下界和进入u的所有弧的流量下界之差，当D(u)>0，则新增一条弧<u,VT>，容量为D(u),当D(u)<0时，则新增一条弧<VS,u>，容量-D(u)，D(u)=0不加弧，原网络的每条弧仍保留，容量改为c(u,v)-b(u,v)。

这个是自己敲的，无优化，主要是为了理解accompany network,下面又给出了优化的代码，效率不是一个数量级的。

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>//这段代码很垃圾 不能拿来当模板 要修改 修改using namespace std;#define min(a,b) (a)>(b)?(b):(a)const int Inf=0x3f3f3f3f;//与memset全设63是等价的 const int maxn=205;int n,m;int source=0 ,sink;struct Arc{    int b,c,f,No_;    //Arc(int bb,int cc,int ff,int nn)b(bb),c(cc),f(ff),No_(nn){};};Arc edge[maxn][maxn];Arc accedge[maxn][maxn];int flag[maxn];int prev[maxn];int alpha[maxn];int queue[maxn];int v,qhead,qrear;int cmp (const void *a , const void *b){    return ((Arc *)a)->No_-((Arc*)b)->No_;}void ford (Arc network[][maxn],int s,int t){    int i,j;    while (1)    {        memset(flag , -1 , sizeof(flag));        memset(prev , -1 , sizeof(prev));        flag[s]=0 ; prev [s]=0 ; alpha[s]=Inf;        qhead=0; queue[0]=s;qrear=1;        while (qhead<qrear && flag[t]==-1)        {            v=queue [qhead]; qhead++;            for (i=s ; i<=t ; ++i)            {                if(flag[i]==-1)                {                    if(network[v][i].c<Inf && network[v][i].f<network[v][i].c)                    {                        flag[i]=0 ; prev[i]=v ;                         alpha[i]=min(alpha[v],network[v][i].c-network[v][i].f);                        queue[qrear]=i;qrear++;                    }                    else if(network[i][v].c<Inf && network[i][v].f> network[i][v].b)                    {                        flag[i]=0 ; prev[i]=-v;                        alpha[i]=min(alpha[v],network[i][v].f-network[i][v].b);                        queue[qrear]=i;qrear++;                    }                }            }            flag[v]=1;        }        if(flag[t]==-1 || alpha[t]==0)break;        int k1=t,k2=fabs(prev[k1]),a=alpha[t];        while (1)        {            if(network[k2][k1].f<Inf)             network[k2][k1].f+=a;            else network[k1][k2].f-=a;            if(k2==s)break;            k1=k2 ; k2=fabs(prev[k2]);        }    }}void accompany ()//构造伴随网络 {    int i,j;    memcpy(accedge,edge , sizeof(edge));    for (i=1 ; i<=n ; i++)     {        int sum1=0 ,sum2=0;        for (j=1 ; j<=n ; ++j)        {            if(accedge[i][j].b!=Inf) sum1+=accedge[i][j].b;            if(accedge[j][i].b!=Inf) sum2+=accedge[j][i].b;        }        if(sum2>sum1)accedge[0][i].c=sum2-sum1,accedge[0][i].b=accedge[0][i].f=0;        else accedge[i][n+1].c=sum1-sum2,accedge[i][n+1].b=accedge[i][n+1].f=0;    }    for (i=1 ; i<=n ; ++i)        for (j=1 ; j<=n ; ++j)if(accedge[i][j].c!=Inf)        {            accedge[i][j].c=accedge[i][j].c-accedge[i][j].b;            accedge[i][j].b=0;        }    ford(accedge,0,n+1);    bool fflag=1;    for (i=0 ; i<=n+1 ; ++i)//若     {        if(accedge[0][i].c!=Inf && accedge[0][i].f!=accedge[0][i].c)fflag=0;    }    if(fflag==0)    {        printf("NO\n");return ;    }     for (i=1 ; i<=n ; i++)    {        for (j=1 ; j<=n ; ++j)        {            if(edge[i][j].c!=Inf)             edge [i][j].f=accedge[i][j].f+edge[i][j].b;        }    }    printf("YES\n");    qsort(edge ,205*205,sizeof(Arc),cmp);//不知道为什么 用sort就不能过    for ( i=0 ; i<m ; ++i ) printf("%d\n",edge[i/m][i%m].f);}int main (){    int u,v,c,b;    int T;    scanf("%d",&T);    while (T--)    {        scanf("%d%d",&n,&m);        memset (edge , 63 , sizeof(edge));        for (int i=0 ; i<m ; ++i)        {            scanf("%d %d %d %d",&u,&v,&b,&c);            edge[u][v].b=b;edge[u][v].c=c;edge[u][v].f=0;edge[u][v].No_=i;        }        accompany();    }    return 0;}

省去了结构体，直接对伴随网络的残留网络进行构造

模板中的res是残留网络，对定义要熟练巩固。

#include <cstdio>#include <string.h>#define min(a,b) ((a)>(b))?(b):(a)using namespace std ;const int maxn=210;const int maxm=40005;const int Inf=0x4fffffff;int cap[maxn][maxn];//原网络int bflow[maxn][maxn];//下限int dist[maxn],gap[maxn];int res[maxn][maxn];//伴随网络的残留网络int x[maxm],y[maxm];int n,m;int dfs (int p , int limit=Inf){    if(p==n)return limit;    for (int i=0 ; i<=n ; ++i)    {        if(dist[p]==dist[i]+1 && res[p][i]>0)        {            int t=dfs(i,min(limit , res[p][i]));            if(t<0)return t;            if(t>0)            {                res[p][i]-=t;//残留网络                res[i][p]+=t;//残留网络的反向弧存的就是当前弧的正向流量                return t;            }        }    }    int tmp=n+1 ;    for (int i=0 ; i<=n ; ++i)        if(res[p][i]>0)            tmp=min(tmp,dist[i]+1);/*printf("dist[p]=%d   dist0=%d   tmp=%d\n",dist[p],dist[0],tmp);printf("gap=");for (int i=0 ; i<n ; i++)printf("%d ",gap[i]);puts("");*/    if(--gap[dist[p]]==0 || dist[0]>n)return -1;    ++gap[dist[p]=tmp];    return 0;}int SAP(){    gap[0]=n+1;    int f = 0 , t=0;    while (~(t=dfs(0))) f+=t;    //printf("%d\n",t);    return f;}void init (){    memset (cap , 0 , sizeof(cap));    memset (res , 0 , sizeof(res));    memset (dist , 0 , sizeof(dist));    memset (gap , 0 , sizeof(gap));    memset (bflow , 0 , sizeof(bflow));}void accompany (){    int i,j,sum1,sum2;    n++;//增加汇点    for (i=1 ; i<n ; ++i)    {        sum1=sum2=0;        for (j=1 ; j<n ; ++j)        {            if(bflow[i][j]>0)sum1+=bflow[i][j];            if(bflow[j][i]>0)sum2+=bflow[j][i];        }        int tmp=sum2-sum1;        tmp>0?(res[0][i]=tmp):res[i][n]=-tmp;    }    SAP();    bool flag=1;    for (i=0 ; i<=n ; ++i)    {        //printf("%d  %d  %d  %d  %d\n",res[0][i] , res[1][i] ,res[2][i] ,res[3][i] ,res[4][i]);        if(res[0][i]!=0)flag=0;    }    if(!flag){printf("NO\n"); return ;}    printf("YES\n");    for (i=0 ; i<m ; ++i)    printf("%d\n",cap[x[i]][y[i]]-res[x[i]][y[i]]);//printf("%d\n",res[y[i]][x[i]]+bflow[x[i]][y[i]]);这个也可以}int main (){    int cas,i,j,u,v,c,b,sum;    //freopen ("in.txt","r",stdin);    //freopen ("out.txt","w",stdout);    scanf("%d",&cas);    while (cas--)    {        init();        scanf("%d%d",&n,&m);        for (i=0 ; i<m ; i++)        {            scanf("%d%d%d%d",x+i,y+i,&b,&c);            cap[x[i]][y[i]]=c;            res[x[i]][y[i]]=c-b;            bflow[x[i]][y[i]]=b;        }        accompany();    }    return 0;}