3698: XWW的难题 有源汇上下界最大流

有源汇上下界最大流，行列建图，下界为 ai,j ，上界为 ai,j+1 ，跑的飞起。
不要忘记判断上取整下取整相同的情况。

#include<iostream>
#include<cstdio>
#include<cstring>
#define inf 1000000007
#define N 205
#define M 30005
using namespace std;
int n,cnt=1,S,T,SS,TT,total,ans;
int head[N],cur[N],dis[N],q[N],du[N];
int next[M],list[M],key[M];
double a[N>>1][N>>1];
int l[N>>1][N>>1];
inline void insert(int x,int y,int z)
{
    next[++cnt]=head[x];
    head[x]=cnt;
    list[cnt]=y;
    key[cnt]=z;
}
inline bool BFS(int S,int T)
{
    int t=0,w=1,x;
    memset(dis,-1,sizeof(dis));
    dis[S]=1; q[1]=S;
    while (t<w)
    {
        x=q[++t];
        for (int i=head[x];i;i=next[i])
            if (key[i]&&dis[list[i]]==-1)
                dis[q[++w]=list[i]]=dis[x]+1;
    }
    return dis[T]!=-1;
}
int find(int x,int flow,int T)
{
    if (x==T) return flow;
    int w,used=0;
    for (int i=cur[x];i;i=next[i])
        if (key[i]&&dis[list[i]]==dis[x]+1)
        {
            w=find(list[i],min(flow-used,key[i]),T);
            key[i]-=w; key[i^1]+=w; used+=w;
            if (key[i]) cur[x]=i;
            if (used==flow) return flow;
        }
    if (!used) dis[x]=-1;
    return used;
}
inline int dinic(int S,int T)
{
    int ans=0;
    while (BFS(S,T))
    {
        for (int i=0;i<=T;i++) cur[i]=head[i];
        ans+=find(S,inf,T);
    }
    return ans;
}
int main()
{
    scanf("%d",&n); n--; S=0; T=n<<1|1; SS=T+1; TT=T+2;
    for (int i=1;i<=n+1;i++)
        for (int j=1;j<=n+1;j++)
            scanf("%lf",&a[i][j]),l[i][j]=(int)a[i][j];
    for (int i=1;i<=n;i++)
    {
        du[i]+=l[i][n+1],du[S]-=l[i][n+1];
        if (a[i][n+1]!=l[i][n+1]) insert(S,i,1),insert(i,S,0);
    }
    for (int i=1;i<=n;i++)
    {
        du[i+n]-=l[n+1][i],du[T]+=l[n+1][i];
        if (a[n+1][i]!=l[n+1][i]) insert(i+n,T,1),insert(T,i+n,0);
    }
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        {
            du[i]-=l[i][j],du[j+n]+=l[i][j];
            if (a[i][j]!=l[i][j]) insert(i,j+n,1),insert(j+n,i,0);
        }
    for (int i=0;i<=T;i++)
        if (du[i]>0) total+=du[i],insert(SS,i,du[i]),insert(i,SS,0);
        else if (du[i]<0) insert(i,TT,-du[i]),insert(TT,i,0);
    insert(T,S,inf); insert(S,T,0);
    ans=dinic(SS,TT);
    if (ans!=total) puts("No");
    else ans=dinic(S,T),printf("%d\n",3*ans);
    return 0;
}