【无源汇上下界可行流】ZOJ2314[Reactor Cooling]题解

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题目概述

有n个点，m根水管，每根水管的流量有上下界，求一种可行流使得每个节点流出的水=流进的水。

解题报告

这道题是经典的无源汇上下界可行流。

示例程序

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=200,maxm=2*(maxn*maxn+maxn),MAXINT=((1<<30)-1)*2+1;int te,n,m,low[maxm+5],num[maxn+5],SS,TT,ful;int E,lnk[maxn+5],cur[maxn+5],dis[maxn+5],que[maxn+5];bool vis[maxn+5];struct Edge{    int cap,flow,nxt,son;Edge() {}    Edge(int a,int b,int c,int d) {cap=a;flow=b;nxt=c;son=d;}};Edge e[maxm+5];void Add(int x,int y,int z){    e[E]=Edge(z,0,lnk[x],y);lnk[x]=E++;    e[E]=Edge(0,0,lnk[y],x);lnk[y]=E++;}bool Bfs(int st,int gl){    memset(vis,0,sizeof(vis));    int Head=0,Tail=0;que[++Tail]=st;dis[st]=0;vis[st]=true;    while (Head!=Tail)    {        int x=que[++Head];        for (int j=lnk[x];~j;j=e[j].nxt)            if (e[j].cap>e[j].flow&&!vis[e[j].son])            {                vis[e[j].son]=true;que[++Tail]=e[j].son;                dis[e[j].son]=dis[x]+1;            }    }    return vis[gl];}int Dfs(int x,int gl,int MIN=MAXINT){    if (!MIN||x==gl) return MIN;    int flow=0,f;    for (int &j=cur[x];~j;j=e[j].nxt)        if (dis[x]+1==dis[e[j].son]&&(f=Dfs(e[j].son,gl,min(MIN,e[j].cap-e[j].flow))))        {            e[j].flow+=f;e[j^1].flow-=f;            flow+=f;MIN-=f;            if (!MIN) break;        }    return flow;}int Dinic(int st,int gl){    int MAX=0;    while (Bfs(st,gl))    {        memcpy(cur,lnk,sizeof(lnk));        MAX+=Dfs(st,gl);    }    return MAX;}int main(){    freopen("program.in","r",stdin);    freopen("program.out","w",stdout);    for (scanf("%d",&te);te;te--)    {        E=0;memset(lnk,255,sizeof(lnk));memset(num,0,sizeof(num));        scanf("%d%d",&n,&m);SS=0;TT=n+1;ful=0;        for (int i=1,x,y,z;i<=m;i++)            scanf("%d%d%d%d",&x,&y,&low[i],&z),num[x]-=low[i],num[y]+=low[i],Add(x,y,z-low[i]);        for (int i=1;i<=n;i++)            if (num[i]>0) Add(SS,i,num[i]),ful+=num[i]; else            if (num[i]<0) Add(i,TT,-num[i]);        if (Dinic(SS,TT)!=ful) printf("NO\n"); else        {            printf("YES\n");            for (int i=0;i<2*m;i+=2)                printf("%d\n",e[i].flow+low[i/2+1]);        }        if (te>1) putchar('\n');    }    return 0;}

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