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;
}