洛谷4013数字梯形

题目：https://www.luogu.org/problemnew/show/P4013

网络流。有无限制就是容量为1还是INF。

发现ek算法难道不能规定源点拥有的流量？

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
typedef long long ll;
const int N=1205,INF=705;
int n,m,head[N],xnt=1,incf[N],pre[N],t;
ll dis[N],ans;
bool in[N];
struct Edge{
    int next,from,to,cap;
    ll w;
    Edge(int n=0,int f=0,int t=0,int c=0,ll w=0):next(n),from(f),to(t),cap(c),w(w) {}
}edge[N<<3],sid[N<<3];
bool spfa()
{
    queue<int> q;
    memset(dis,-2,sizeof dis);
    memset(pre,0,sizeof pre);
    q.push(0);in[0]=1;dis[0]=0;incf[0]=INF;
    while(q.size())
    {
        int k=q.front();q.pop();in[k]=0;
//        printf("k=%d
",k);
        for(int i=head[k],v;i;i=sid[i].next)
        {
//            printf(" (to=%d)
",sid[i].to);
            if(sid[i].cap&&dis[k]+sid[i].w>dis[v=sid[i].to])
            {
//                printf(" v=%d
",v);
                dis[v]=dis[k]+sid[i].w;
                incf[v]=min(incf[k],sid[i].cap);
                pre[v]=i;
                if(!in[v])q.push(v),in[v]=1;
            }
        }
    }
    return pre[t];
}
void ek()
{
//    printf("(%d)",incf[t]);
    ans+=dis[t]*incf[t];
    for(int k=pre[t];k;k=pre[sid[k].from])
    {
        sid[k].cap-=incf[t];
        sid[k^1].cap+=incf[t];
    }
}
void solve()
{
    while(spfa())
        ek();
    printf("%lld
",ans);
    ans=0;
}
int main()
{
    scanf("%d%d",&m,&n);
    int num=(m+(m+n-1))*n/2;
    ll x,bh=0;t=2*num+1;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m+i-1;j++)
        {
            int k=bh+j,u=k-(m+i-2);
            scanf("%lld",&x);
//            printf("(k=%d u=%d)
",k,u);
            edge[++xnt]=Edge(head[k],k,k+num,1,x);head[k]=xnt;
            edge[++xnt]=Edge(head[k+num],k+num,k,0,-x);head[k+num]=xnt;
            if(i==1)
            {
                edge[++xnt]=Edge(head[0],0,k,1,0);head[0]=xnt;
                edge[++xnt]=Edge(head[k],k,0,0,0);head[k]=xnt;
                continue;
            }
            if(i==n)
            {
                edge[++xnt]=Edge(head[k+num],k+num,t,INF,0);head[k+num]=xnt;
                edge[++xnt]=Edge(head[t],t,k+num,0,0);head[t]=xnt;
            }
            if(j!=m+i-1)
            {
                edge[++xnt]=Edge(head[u+num],u+num,k,1,0);head[u+num]=xnt;
                edge[++xnt]=Edge(head[k],k,u+num,0,0);head[k]=xnt;
            }
            if(j!=1)
            {
                edge[++xnt]=Edge(head[u-1+num],u-1+num,k,1,0);head[u-1+num]=xnt;
                edge[++xnt]=Edge(head[k],k,u-1+num,0,0);head[k]=xnt;
            }
        }
        bh+=m+i-1;
    }
    memcpy(sid,edge,sizeof edge);
    solve();
    memcpy(sid,edge,sizeof edge);
    for(int i=2;i<=xnt;i++)
        if(sid[i].w&&sid[i].cap)sid[i].cap=INF;
    solve();
    memcpy(sid,edge,sizeof edge);
    for(int i=2;i<=xnt;i++)
        if(sid[i].cap&&sid[i].to>m)sid[i].cap=INF;
    solve();
    return 0;
}