[POJ2396]Budget（有源汇有上下界的可行流）

来源：互联网发布：什么是数据编码技术编辑：程序博客网时间：2024/05/22 15:11

题目描述

传送门

题解

和无源汇有上下界的可行流类似，只需要加一条边t->s，限制为[0,inf]就可以了，相当于是让源点和汇点也满足流量平衡
然后再建立附加源汇ss,tt，其余的就和无源汇的一样了

这道题的话，把每一行看成一个点xi，每一列看成一个点yi，那么首先s->xi,[sumxi,sumxi]，yi->t,[sumyi,sumyi]
对于某一个点(i,j)的限制[l,r]，xi->yj,[l,r]

代码

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<queue>using namespace std;#define N 10005#define inf 1000000000char opt;int T,n,m,k,x,y,z,s,t,ss,tt,maxflow,in,out;int l[205][205],r[205][205],d[N],pipe[205][205];int tot,point[N],nxt[N],v[N],remain[N];int deep[N],last[N],cur[N],num[N];queue <int> q;void clear(){    for (int i=1;i<=n;++i)        for (int j=1;j<=m;++j)            l[i][j]=0,r[i][j]=inf;    maxflow=in=out=0;    memset(d,0,sizeof(d));    tot=-1;memset(point,-1,sizeof(point));    memset(num,0,sizeof(num));}void addedge(int x,int y,int cap){    ++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap;    ++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0;}void bfs(int t){    for (int i=1;i<=t;++i) deep[i]=t;    deep[t]=0;    for (int i=1;i<=t;++i) cur[i]=point[i];    while (!q.empty()) q.pop();    q.push(t);    while (!q.empty())    {        int now=q.front();q.pop();        for (int i=point[now];i!=-1;i=nxt[i])            if (deep[v[i]]==t&&remain[i^1])            {                deep[v[i]]=deep[now]+1;                q.push(v[i]);            }    }}int addflow(int s,int t){    int now=t,ans=inf;    while (now!=s)    {        ans=min(ans,remain[last[now]]);        now=v[last[now]^1];    }    now=t;    while (now!=s)    {        remain[last[now]]-=ans;        remain[last[now]^1]+=ans;        now=v[last[now]^1];    }    return ans;}void isap(int s,int t){    bfs(t);    for (int i=1;i<=t;++i) ++num[deep[i]];    int now=s;    while (deep[s]<t)    {        if (now==t)        {            maxflow+=addflow(s,t);            now=s;        }        bool has_find=false;        for (int i=cur[now];i!=-1;i=nxt[i])            if (deep[v[i]]+1==deep[now]&&remain[i])            {                has_find=true;                cur[now]=i;                last[v[i]]=i;                now=v[i];                break;            }        if (!has_find)        {            int minn=t-1;            for (int i=point[now];i!=-1;i=nxt[i])                if (remain[i]) minn=min(minn,deep[v[i]]);            if (!(--num[deep[now]])) break;            ++num[deep[now]=minn+1];            cur[now]=point[now];            if (now!=s) now=v[last[now]^1];        }    }}int main(){    scanf("%d",&T);    while (T--)    {        scanf("%d%d",&n,&m);        clear();        s=n+m+1,t=s+1,ss=t+1,tt=ss+1;        for (int i=1;i<=n;++i) scanf("%d",&x),d[s]-=x,d[i]+=x;        for (int i=1;i<=m;++i) scanf("%d",&x),d[n+i]-=x,d[t]+=x;        scanf("%d",&k);        while (k--)        {            scanf("%d %d %c %d",&x,&y,&opt,&z);            if (!x&&!y)                for (int i=1;i<=n;++i)                    for (int j=1;j<=m;++j)                    {                        if (opt=='=') l[i][j]=r[i][j]=z;                        if (opt=='>') l[i][j]=max(l[i][j],z+1);                        if (opt=='<') r[i][j]=min(r[i][j],z-1);                    }            else if (!x)                for (int i=1;i<=n;++i)                {                    if (opt=='=') l[i][y]=r[i][y]=z;                    if (opt=='>') l[i][y]=max(l[i][y],z+1);                    if (opt=='<') r[i][y]=min(r[i][y],z-1);                }            else if (!y)                for (int i=1;i<=m;++i)                {                    if (opt=='=') l[x][i]=r[x][i]=z;                    if (opt=='>') l[x][i]=max(l[x][i],z+1);                    if (opt=='<') r[x][i]=min(r[x][i],z-1);                }            else            {                if (opt=='=') l[x][y]=r[x][y]=z;                if (opt=='>') l[x][y]=max(l[x][y],z+1);                if (opt=='<') r[x][y]=min(r[x][y],z-1);            }        }        addedge(t,s,inf);        for (int i=1;i<=n;++i)            for (int j=1;j<=m;++j)            {                addedge(i,n+j,r[i][j]-l[i][j]),pipe[i][j]=tot;                d[i]-=l[i][j],d[n+j]+=l[i][j];            }        for (int i=1;i<=t;++i)        {            if (d[i]>0) addedge(ss,i,d[i]),in+=d[i];            if (d[i]<0) addedge(i,tt,-d[i]),out-=d[i];        }        if (in!=out) {puts("IMPOSSIBLE");puts("");continue;}        isap(ss,tt);        if (maxflow==in)        {            for (int i=1;i<=n;++i)                for (int j=1;j<=m;++j)                    printf("%d%c",remain[pipe[i][j]]+l[i][j]," \n"[j==m]);        }        else puts("IMPOSSIBLE");        puts("");    }}

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