[BZOJ3698]XWW的难题(有源汇有上下界的最大流)

题目描述

传送门

题解

最大流和可行流的做法的区别:先ss->tt做一遍最大流,判断是否可行;然后将t->s,inf这条边去掉,再做一遍s->t的最大流,即为答案
这道题原图的建图方法是:
对于每一行i,s->i,[a(i,n),a(i,n)+1]
对于每一列j,j->t,[a(n,j),a(n,j)+1]
对于每一个点(i,j),i->j,[a(i,j),a(i,j)+1]
然后再按照有源汇有上下界对这个图进行改造即可

代码

#include
#include
#include
#include
#include
#include
using namespace std;
#define N 100005
#define inf 1000000000

int n,s,t,ss,tt,maxflow,in,out,ans;
double a[105][105];
int l[105][105],r[105][105],p[105][105];
int tot,point[N],nxt[N],v[N],remain[N];
int d[N],deep[N],last[N],cur[N],num[N];
queue <int> q;

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]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()
{
    tot=-1;memset(point,-1,sizeof(point));
    scanf("%d",&n);
    for (int i=1;i<=n;++i)
        for (int j=1;j<=n;++j)
        {
            scanf("%lf",&a[i][j]);
            l[i][j]=floor(a[i][j]);
            r[i][j]=ceil(a[i][j]);
        }
    s=n+n+1,t=s+1,ss=t+1,tt=ss+1;
    for (int i=1;ifor (int i=1;ifor (int j=1;jfor (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];
    }
    addedge(t,s,inf);
    if (in!=out) {puts("NO");return 0;}
    isap(ss,tt);
    if (maxflow!=in) {puts("NO");return 0;}
    remain[tot]=remain[tot^1]=0;
    isap(s,t);

    for (int i=1;ifor (int j=1;jprintf("%d\n",ans*3);

    return 0;
}

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