题目链接：戳我
拆点。第一种情况流量为1，第二种情况同一个点之间连inf，第三种情况除了第一行连1，其他都inf。
做完了。注意空间还是要开够，而且数值还有负数。
跑最大费用最大流即可。
代码如下：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#define MAXN 500010
using namespace std;
int n,m,S=0,T,cnt,c,f,t=1,tot;
int head[MAXN],val[30][41],pre_v[MAXN],pre_e[MAXN],dis[MAXN],done[MAXN],id[30][41];
struct Edge{int nxt,to,dis,cost;}edge[MAXN<<1];
inline void add(int from,int to,int dis,int cost)
{
edge[++t].nxt=head[from],edge[t].to=to,edge[t].dis=dis,edge[t].cost=cost,head[from]=t;
edge[++t].nxt=head[to],edge[t].to=from,edge[t].dis=0,edge[t].cost=-cost,head[to]=t;
}
inline bool spfa()
{
queue<int>q;
memset(dis,0x3f,sizeof(dis));
memset(done,0,sizeof(done));
q.push(S);dis[S]=0;done[S]=1;
while(!q.empty())
{
int u=q.front();q.pop();done[u]=0;
for(int i=head[u];i;i=edge[i].nxt)
{
int v=edge[i].to;
if(edge[i].dis&&dis[u]+edge[i].cost<dis[v])
{
dis[v]=dis[u]+edge[i].cost;
pre_e[v]=i,pre_v[v]=u;
if(!done[v])
q.push(v),done[v]=1;
}
}
}
if(dis[T]==0x3f3f3f3f) return false;
int flow=0x3f3f3f3f;
for(int i=T;i!=S;i=pre_v[i]) flow=min(flow,edge[pre_e[i]].dis);
for(int i=T;i!=S;i=pre_v[i]) edge[pre_e[i]].dis-=flow,edge[pre_e[i]^1].dis+=flow;
f+=flow;
c+=flow*dis[T];
return true;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("ce.in","r",stdin);
#endif
scanf("%d%d",&m,&n);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i+m-1;j++)
{
scanf("%d",&val[i][j]);
id[i][j]=++cnt;
}
}
tot=cnt;
T=cnt*2+1;
//printf("S=%d T=%d tot=%d\n",S,T,tot);
//subtask1
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i+m-1;j++)
{
add(id[i][j],id[i][j]+tot,1,0);
if(i==1) add(S,id[i][j],1,-val[i][j]);
if(i==n) add(id[i][j]+tot,T,1,0);
if(i!=n)
{
add(id[i][j]+tot,id[i+1][j],1,-val[i+1][j]);
add(id[i][j]+tot,id[i+1][j+1],1,-val[i+1][j+1]);
}
}
}
while(spfa());
printf("%d\n",-c);
//subtask2
c=f=0;t=1;
memset(head,0,sizeof(head));
memset(pre_e,0,sizeof(pre_e));
memset(pre_v,0,sizeof(pre_v));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i+m-1;j++)
{
add(id[i][j],id[i][j]+tot,0x3f3f3f3f,0);
if(i==1) add(S,id[i][j],1,-val[i][j]);
if(i==n) add(id[i][j]+tot,T,0x3f3f3f3f,0);
if(i!=n)
{
add(id[i][j]+tot,id[i+1][j],1,-val[i+1][j]);
add(id[i][j]+tot,id[i+1][j+1],1,-val[i+1][j+1]);
}
}
}
while(spfa());
printf("%d\n",-c);
//subtask3
c=f=0;t=1;
memset(head,0,sizeof(head));
memset(pre_e,0,sizeof(pre_e));
memset(pre_v,0,sizeof(pre_v));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i+m-1;j++)
{
add(id[i][j],id[i][j]+tot,0x3f3f3f3f,-val[i][j]);
if(i==1) add(S,id[i][j],1,0);
if(i==n) add(id[i][j]+tot,T,0x3f3f3f3f,0);
if(i!=n)
{
add(id[i][j]+tot,id[i+1][j],0x3f3f3f3f,0);
add(id[i][j]+tot,id[i+1][j+1],0x3f3f3f3f,0);
}
}
}
while(spfa());
printf("%d\n",-c);
return 0;
}
