[BZOJ3698]XWW的难题（有源汇有上下界的最大流）

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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<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<queue>using namespace std;#define N 100005#define inf 1000000000int 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]<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(){    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;i<n;++i)    {        addedge(s,i,r[i][n]-l[i][n]);        d[s]-=l[i][n],d[i]+=l[i][n];        addedge(n+i,t,r[n][i]-l[n][i]);        d[n+i]-=l[n][i],d[t]+=l[n][i];    }    for (int i=1;i<n;++i)        for (int j=1;j<n;++j)        {            addedge(i,n+j,r[i][j]-l[i][j]);            p[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];    }    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;i<n;++i)        for (int j=1;j<n;++j)            ans+=remain[p[i][j]]+l[i][j];    printf("%d\n",ans*3);    return 0;}

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