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; }