POJ2396_Budget

题意为给一个矩形数字阵，给出一些限制条件，包括每行和每列的和，还有一些位置的数值范围，求出满足情况的一个。

首先建图，源点->行和->列和->汇点，显然，行和列之间的边为那个数字的大小，只要我们能够找到一个满足大小条件的，且使的两边的和满流的流量方案就可以了。

由于存在下界（上界其实就是边的容量），根据图的特殊性，我们可以先在那边的相连的两条边都减去这个下界，这样就变成了一条只有上界的边了。

 

 

召唤代码君：

 

 

#include <iostream>
#include <cstring>
#include <cstdio>
#define maxn 1022
#define maxm 844442
typedef long long ll;
using namespace std;

const ll inf=~0U>>1;
ll to[maxm],next[maxm],c[maxm],first[maxn],edge;
ll fmin[maxn][maxn],fmax[maxn][maxn],sr[maxn],sc[maxn];
ll d[maxn],tag[maxn],TAG=222;
ll Q[maxm],bot,top;
ll ans[maxn][maxn];
bool can[maxn];
ll n,m,s,t,T,R,sumr,sumc;

void _init()
{
    s=0,t=n+m+1,edge=-1,sumr=sumc=0;
    for (ll i=s; i<=t; i++) first[i]=-1;
    for (ll i=1; i<=n; i++) sr[i]=0;
    for (ll i=1; i<=m; i++) sc[i]=0;
    for (ll i=1; i<=n; i++)
        for (ll j=1; j<=m; j++) ans[i][j]=0,fmin[i][j]=0,fmax[i][j]=inf;
}

void minsize(ll x,ll y,ll dn,ll up)
{
    fmin[x][y]=max(fmin[x][y],dn);
    fmax[x][y]=min(fmax[x][y],up);
}

bool check()
{
    for (ll i=1; i<=n; i++)
        for (ll j=1; j<=m; j++)
            {
                if (fmax[i][j]<0) return false;
                sr[i]-=fmin[i][j],sc[j]-=fmin[i][j];
                sumr-=fmin[i][j],sumc-=fmin[i][j];
                fmax[i][j]-=fmin[i][j];
                if (fmax[i][j]<0 || sr[i]<0 || sc[j]<0) return false;
            }
    return sumr==sumc;
}

void addedge(ll U,ll V,ll W)
{
    edge++;
    to[edge]=V,c[edge]=W,next[edge]=first[U],first[U]=edge;
    edge++;
    to[edge]=U,c[edge]=0,next[edge]=first[V],first[V]=edge;
}

bool bfs()
{
    Q[bot=top=1]=t,tag[t]=++TAG,d[t]=0,can[t]=false;
    while (bot<=top)
    {
        ll cur=Q[bot++];
        for (ll i=first[cur]; i!=-1; i=next[i])
            if (c[i^1] && tag[to[i]]!=TAG)
            {
                tag[to[i]]=TAG,d[to[i]]=d[cur]+1;
                can[to[i]]=false,Q[++top]=to[i];
                if (to[i]==s) return true;
            }
    }
    return false;
}

ll dfs(ll cur,ll num)
{
    if (cur==t) return num;
    ll tmp=num,k;
    for (ll i=first[cur]; i!=-1; i=next[i])
        if (c[i] && d[to[i]]==d[cur]-1 && tag[to[i]]==TAG && !can[to[i]])
        {
            k=dfs(to[i],min(c[i],num));
            if (k) num-=k,c[i]-=k,c[i^1]+=k;
            if (!num) break;
        }
    if (num) can[cur]=true;
    return tmp-num;
}

ll maxflow()
{
    ll tot=0;
    while (bfs()) tot+=dfs(s,~0U>>1);
    return tot;
}

int main()
{
    char S[3];
    ll x,y,z,cas=0,up,dn;
    scanf("%I64d",&T);
    while (T--)
    {
        scanf("%I64d%I64d",&n,&m);
        _init();
        for (ll i=1; i<=n; i++) scanf("%I64d",&sr[i]),sumr+=sr[i];
        for (ll i=1; i<=m; i++) scanf("%I64d",&sc[i]),sumc+=sc[i];
        scanf("%I64d",&R);
        while (R--)
        {
            scanf("%I64d%I64d%s%I64d",&x,&y,S,&z);
            if (S[0]=='=') up=z,dn=max(0LL,z);
                else if (S[0]=='>') up=inf,dn=max(0LL,z+1);
                    else up=z-1,dn=0;
            if (x==0 && y==0)
            {
                for (ll i=1; i<=n; i++)
                    for (ll j=1; j<=m; j++) minsize(i,j,dn,up);
            }
            else if (x==0)
            {
                for (ll i=1; i<=n; i++) minsize(i,y,dn,up);
            }
            else if (y==0)
            {
                for (ll j=1; j<=m; j++) minsize(x,j,dn,up);
            }
            else minsize(x,y,dn,up);
        }
        if (cas++) puts("");
        if (!check())
        {
            puts("IMPOSSIBLE");
            continue;
        }
        for (ll i=1; i<=n; i++) addedge(s,i,sr[i]);
        for (ll j=1; j<=m; j++) addedge(n+j,t,sc[j]);
        for (ll i=1; i<=n; i++)
            for (ll j=1; j<=m; j++)
                if (fmax[i][j]>0) addedge(i,n+j,fmax[i][j]);
        /*        
        for (int i=1; i<=n; i++)
        {
            cout<<" hehe : ";
            for (int j=1; j<=m; j++) cout<<fmin[i][j]<<"("<<fmax[i][j]<<") ... ";
            cout<<endl;
        }
        */
        if (maxflow()!=sumr)
        {
            puts("IMPOSSIBLE");
            continue;
        }
        for (ll i=n+n+m+m; i<=edge; i+=2)
        {
            x=to[i+1],y=to[i]-n;
            ans[x][y]=c[i+1];
        }
        for (ll i=1; i<=n; i++)
        {
            printf("%I64d",ans[i][1]+fmin[i][1]);
            for (ll j=2; j<=m; j++) printf(" %I64d",ans[i][j]+fmin[i][j]);
            printf("\n");
        }
    } 
    return 0;
}

 

转载于:https://www.cnblogs.com/lochan/p/3870725.html