题目链接: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=3894
(luogu) https://www.luogu.org/problemnew/show/P4313
题解:
做法很简单，就是最小割，\(S\)集属于文科，\(T\)集属于理科，对于每个点\(i\), 起点\(S\)向\(i\)连\(a_i\)(文科收益/理科代价)，\(i\)向终点\(T\)连\(b_i\) (理科收益/文科代价)，对于每一个点\(i\)再新建两点\(i_a\)(同文点)和\(i_b\)(同理点)，\(S\)向\(i_a\)连边\(aa_i\)(同文收益)，\(i_b\)向\(T\)连\(bb_i\)(同理收益)，中间对于\(i\)和\(i\)座位相连的每个点，从\(i_a\)向该点连边，从该点向\(i_b\)连边，边权均为\(+\inf\).
我的错误做法: 如果同文同理建成同一个点，和座位相连的每个点连双向边，那么这是错的，如果连单向边也是错的。因为建两个点实际上可以保证如果\(i_a\)属于\(S\)集则它连向的人都选文，如果\(i_b\)属于\(T\)集则连向它的人都选理，如果它们与\(S,T\)之间的边都被割掉了，则它们对这些人没有任何限制，这些人仍是独立的。但如果同文同理建成同一个点连双向边，那么这些点之间构成强连通分量，相当于默认所有人必须在同一集合，这是最离谱的做法我居然能想出来。如果连单向边呢，比如从新点往这几个人连边，从\(S\)往新点连边，从新点往\(T\)连边，那么相当于规定“如果新点属于\(S\)则这些人全属于\(S\), 如果新点属于\(T\)则对这些人没有要求”。总之，从\(i\)往\(j\)连边\(\inf\)则相当于如果\(i\in S\)则\(j\in S\), 但是如果\(i\in T\)则对\(j\)没有要求；如果\(j\in T\)则\(i\in T\)，而如果\(j\in S\)则没有要求对\(i\)没有要求(这两句话是等价的)。\(i\)和\(j\)之间连双向\(\inf\)边则相当于强制两点在同一集合中。
代码
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cassert>
using namespace std;
const int INF = 1e8;
namespace MaxFlow
{
const int N = 3e4+2;
const int M = 14e4;
struct Edge
{
int v,w,nxt,rev;
} e[(M<<1)+3];
int fe[N+3];
int te[N+3];
int que[N+3];
int dep[N+3];
int n,en,s,t;
void addedge(int u,int v,int w)
{
en++; e[en].v = v; e[en].w = w;
e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;
en++; e[en].v = u; e[en].w = 0;
e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;
}
bool bfs()
{
for(int i=1; i<=n; i++) dep[i] = 0;
int head = 1,tail = 1; que[tail] = s; dep[s] = 1;
while(head<=tail)
{
int u = que[head]; head++;
for(int i=fe[u]; i; i=e[i].nxt)
{
if(dep[e[i].v]==0 && e[i].w>0)
{
dep[e[i].v] = dep[u]+1;
tail++; que[tail] = e[i].v;
}
}
}
return dep[t]!=0;
}
int dfs(int u,int cur)
{
if(u==t) {return cur;}
int rst = cur;
for(int i=te[u]; i; i=e[i].nxt)
{
if(dep[e[i].v]==dep[u]+1 && e[i].w>0 && rst>0)
{
int flow = dfs(e[i].v,min(rst,e[i].w));
if(flow>0)
{
rst -= flow; e[i].w -= flow; e[e[i].rev].w += flow;
if(e[i].w>0) {te[u] = i;}
if(rst==0) return cur;
}
}
}
if(cur==rst) dep[u] = 0;
return cur-rst;
}
int dinic(int _n,int _s,int _t)
{
int ret = 0;
n = _n,s = _s,t = _t;
while(bfs())
{
for(int i=1; i<=n; i++) te[i] = fe[i];
ret += dfs(s,INF);
}
return ret;
}
}
using MaxFlow::addedge;
using MaxFlow::dinic;
const int N = 100;
int a[N+3][N+3],b[N+3][N+3],aa[N+3][N+3],bb[N+3][N+3];
int n,m;
int getid(int x,int y) {return (x-1)*m+y+2;}
int main()
{
scanf("%d%d",&n,&m); int ans = 0;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
scanf("%d",&a[i][j]); ans += a[i][j];
}
}
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
scanf("%d",&b[i][j]); ans += b[i][j];
}
}
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
scanf("%d",&aa[i][j]); ans += aa[i][j];
}
}
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
scanf("%d",&bb[i][j]); ans += bb[i][j];
}
}
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
int x = getid(i,j);
addedge(1,x,a[i][j]);
addedge(x,2,b[i][j]);
addedge(1,x+n*m,aa[i][j]);
addedge(x+n*m*2,2,bb[i][j]);
addedge(x+n*m,x,INF);
addedge(x,x+n*m*2,INF);
if(i>1)
{
addedge(x+n*m,getid(i-1,j),INF);
addedge(getid(i-1,j),x+n*m*2,INF);
}
if(j>1)
{
addedge(x+n*m,getid(i,j-1),INF);
addedge(getid(i,j-1),x+n*m*2,INF);
}
if(i<n)
{
addedge(x+n*m,getid(i+1,j),INF);
addedge(getid(i+1,j),x+n*m*2,INF);
}
if(j<m)
{
addedge(x+n*m,getid(i,j+1),INF);
addedge(getid(i,j+1),x+n*m*2,INF);
}
}
}
int tmp = dinic(n*m*3+2,1,2);
ans -= tmp;
printf("%d\n",ans);
return 0;
}
