题目链接
(Luogu) https://www.luogu.org/problem/P3756
(BZOJ) http://lydsy.com/JudgeOnline/problem.php?id=4823
题解
有点神仙的最小割题。
考虑题目里的图形，如果我们用四种颜色对棋盘进行染色，奇数行依次染\(0,1,2,3,0,1,2,3...\), 偶数行依次染\(3,2,1,0,3,2,1,0...\)则条件可以转化为不能出现相连的\(4\)个颜色互不相同的块。
那么可以建一个四层的图，对于每条两侧都有关键点的特殊边，按照\(S\rightarrow 3\rightarrow 0\rightarrow 1\rightarrow 2\rightarrow T\)的顺序连边，其中\(S\rightarrow 3\)连\(3\)色点的点权，\(0\rightarrow 1\)连两个关键点权值的最小值，\(2\rightarrow T\)连\(2\)色点的点权。不出现相连的四个颜色互不相同的块等价于不存在从\(S\)到\(T\)的路径。
然后跑最小割即可。
因为是分层图，所以dinic跑得很快(复杂度应该是\(O(n\sqrt n)\))，可以通过此题。
代码
#include<bits/stdc++.h>
#define llong long long
using namespace std;
const int INF = 1e9;
namespace NetFlow
{
const int N = 1e5+2;
const int M = 8e5;
struct Edge
{
int v,w,nxt,rev;
} e[(M<<1)+3];
int fe[N+3];
int te[N+3];
int dep[N+3];
int que[N+3];
int n,en,s,t;
void addedge(int u,int v,int w)
{
// printf("addedge %d %d %d\n",u,v,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[1] = s; dep[s] = 1;
while(head<=tail)
{
int u = que[head]; head++;
for(int i=fe[u]; i; i=e[i].nxt)
{
int v = e[i].v;
if(e[i].w>0 && dep[v]==0)
{
dep[v] = dep[u]+1;
if(v==t)return true;
tail++; que[tail] = v;
}
}
}
return false;
}
int dfs(int u,int cur)
{
if(u==t||cur==0) {return cur;}
int rst = cur;
for(int &i=te[u]; i; i=e[i].nxt)
{
int v = e[i].v;
if(e[i].w>0 && rst>0 && dep[v]==dep[u]+1)
{
int flow = dfs(v,min(rst,e[i].w));
if(flow>0)
{
e[i].w -= flow;
rst -= flow;
e[e[i].rev].w += flow;
if(rst==0) {return cur;}
}
}
}
if(rst==cur) {dep[u] = -2;}
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];
memcpy(te,fe,sizeof(int)*(n+1));
ret += dfs(s,INF);
}
return ret;
}
}
using NetFlow::addedge;
using NetFlow::dinic;
const int N = 1e5;
struct Point
{
int x,y,w;
} a[N+3];
map<int,int> mp[N+3];
int id[N+3],clr[N+3];
int n,nx,ny;
int getclr(int x,int y)
{
if(y&1) {return (x-1)&3;}
else {return 3-((x-1)&3);}
}
int main()
{
scanf("%d%d%d",&nx,&ny,&n);
for(int i=1; i<=n; i++)
{
scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].w);
mp[a[i].x][a[i].y] = i;
}
for(int i=1; i<=n; i++)
{
int x = a[i].x,y = a[i].y,clr = getclr(x,y);
if(((x+(y<<1))&3)==3 && mp[x+1].count(y))
{
int j = mp[x+1][y],w = min(a[i].w,a[j].w);
if(clr==0)
{
int k = mp[x-1][y]; if(k) {addedge(k+2,i+2,INF);}
k = mp[x][y-1]; if(k) {addedge(k+2,i+2,INF);}
k = mp[x][y+1]; if(k) {addedge(k+2,i+2,INF);}
k = mp[x+2][y]; if(k) {addedge(j+2,k+2,INF);}
k = mp[x+1][y+1]; if(k) {addedge(j+2,k+2,INF);}
k = mp[x+1][y-1]; if(k) {addedge(j+2,k+2,INF);}
addedge(i+2,j+2,w);
}
else if(clr==1)
{
int k = mp[x+2][y]; if(k) {addedge(k+2,j+2,INF);}
k = mp[x+1][y+1]; if(k) {addedge(k+2,j+2,INF);}
k = mp[x+1][y-1]; if(k) {addedge(k+2,j+2,INF);}
k = mp[x-1][y]; if(k) {addedge(i+2,k+2,INF);}
k = mp[x][y-1]; if(k) {addedge(i+2,k+2,INF);}
k = mp[x][y+1]; if(k) {addedge(i+2,k+2,INF);}
addedge(j+2,i+2,w);
}
}
else if(clr==3)
{
addedge(1,i+2,a[i].w);
}
else if(clr==2)
{
addedge(i+2,2,a[i].w);
}
}
int ans = dinic(n+2,1,2);
printf("%d\n",ans);
return 0;
}
