题目链接: https://www.lydsy.com/JudgeOnline/problem.php?id=2127
题解: 这道题就是传说中的“解方程”法。(貌似也有类似于BZOJ 3894的做法,但是边数比较多。)
以下设(A_i)为(i)选文的收益,(B_i)为(i)选理的收益,(AA_{i,j})表示(i,j)同文的收益,(BB_{i,j})表示(i,j)同理的收益。
首先对于每个点(i), 从(S)向(i)连(A_i),从(i)向(T)连(B_i)这个没有问题。
然后考虑处理同文同理的代价。
考虑(S,T,i,j)共(4)个点之间可以连(6)条边,设(S)到(i), (i)到(T), (S)到(j), (j)到(T), (i)到(j), (j)到(i)的容量分别为(a,b,c,d,e,f).
枚举(i,j)选文理的四种情况可以列出四个方程。
若(iin S, jin S), 则割掉的边是(b)和(d), 代价是两人不可同理(science)(注意单人文理的代价已经在刚才算过了,所以不要再算!)可得(b+d=BB_{i,j}), 同理(reason)可得(a+c=AA_{i,j}).
若(iin S, jin T), 则割掉的边是(b,c,e) (一定注意没有(f)), 代价是二人不可同文或同理(science), 可得(b+c+e=AA_{i,j}+BB_{i,j}), 同理(reason)可得(a+d+f=AA_{i,j}+BB_{i,j}).
这样我们列出了(4)个方程,给(6)个变量复制绰绰有余,可以随便取值。但是注意也不能太随便,比如不能出负数等等。一种比较好的取法是: $$a=c=frac{AA_{i,j}}{2},b=d=frac{BB_{i,j}}{2},e=f=frac{AA_{i,j}+BB_{i,j}}{2}$$
最后合并起点终点均相同的边来减少边数,除以(2)可以先乘(2)再把答案除以(2)处理,避免出现小数。
好神仙啊……
代码
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF = 4e8;
namespace MaxFlow
{
const int N = 1e4+2;
const int M = 6e4;
struct Edge
{
int v,w,nxt,rev;
} e[(M<<1)+3];
int fe[N+3],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 && rst>0 && e[i].w>0)
{
int flow = dfs(e[i].v,min(rst,e[i].w));
if(flow>0)
{
e[i].w -= flow; e[e[i].rev].w += flow; rst -= flow;
if(e[i].w>0) te[u] = i;
if(rst==0) return cur;
}
}
}
if(rst==cur) dep[u] = 0;
return cur-rst;
}
int dinic(int _n,int _s,int _t)
{
n = _n,s = _s,t = _t;
int ret = 0;
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];
int b[N+3][N+3];
int aa1[N+3][N+3];
int aa2[N+3][N+3];
int bb1[N+3][N+3];
int bb2[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",&aa1[i][j]),ans += aa1[i][j];
}
for(int i=1; i<n; i++)
{
for(int j=1; j<=m; j++) scanf("%d",&bb1[i][j]),ans += bb1[i][j];
}
for(int i=1; i<=n; i++)
{
for(int j=1; j<m; j++) scanf("%d",&aa2[i][j]),ans += aa2[i][j];
}
for(int i=1; i<=n; i++)
{
for(int j=1; j<m; j++) scanf("%d",&bb2[i][j]),ans += bb2[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]<<1)+aa1[i][j]+aa1[i-1][j]+aa2[i][j]+aa2[i][j-1]);
addedge(x,2,(b[i][j]<<1)+bb1[i][j]+bb1[i-1][j]+bb2[i][j]+bb2[i][j-1]);
if(i<n)
{
int y = getid(i+1,j);
addedge(x,y,aa1[i][j]+bb1[i][j]);
addedge(y,x,aa1[i][j]+bb1[i][j]);
}
if(j<m)
{
int y = getid(i,j+1);
addedge(x,y,aa2[i][j]+bb2[i][j]);
addedge(y,x,aa2[i][j]+bb2[i][j]);
}
}
}
int tmp = dinic(n*m+2,1,2);
ans -= (tmp>>1);
printf("%d
",ans);
return 0;
}