标签:最大 class namespace void algorithm cond 拓扑 main putc
图中的保护关系就类似于最大权闭合子图。即你想杀x,你就一定要杀掉保护x的点,那么把x向保护它的点连边。那么题目就转化成了最大权闭合子图的问题。
但是这个图有点特殊啊。。。 考虑有环的情况,显然这个环以及指向这个环的点都不能选。
所以还要把这个图的反图进行一遍拓扑排序,这样忽略掉了这些点了。。。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-9 # define MOD 1024523 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar(‘-‘); a=-a;} if(a>=10) Out(a/10); putchar(a%10+‘0‘); } const int N=605; //Code begin... struct Edge{int p, next, w;}edge[N*N]; int head[N], cnt=2, cost[N], dee[N], n, m, s, t, vis[N]; bool mark[N]; queue<int>Q; vector<PII> E[N]; void add_edge(int u, int v, int w){ edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++; edge[cnt].p=u; edge[cnt].w=0; edge[cnt].next=head[v]; head[v]=cnt++; } int bfs(){ int i, v; mem(vis,-1); vis[s]=0; Q.push(s); while (!Q.empty()) { v=Q.front(); Q.pop(); for (i=head[v]; i; i=edge[i].next) { if (edge[i].w>0 && vis[edge[i].p]==-1) { vis[edge[i].p]=vis[v] + 1; Q.push(edge[i].p); } } } return vis[t]!=-1; } int dfs(int x, int low){ int i, a, temp=low; if (x==t) return low; for (i=head[x]; i; i=edge[i].next) { if (edge[i].w>0 && vis[edge[i].p]==vis[x]+1){ a=dfs(edge[i].p,min(edge[i].w,temp)); temp-=a; edge[i].w-=a; edge[i^1].w += a; if (temp==0) break; } } if (temp==low) vis[x]=-1; return low-temp; } void Topsort(){ FO(i,0,n*m) if (!dee[i]) Q.push(i), mark[i]=true; while (!Q.empty()) { int u=Q.front(); Q.pop(); FO(i,0,E[u].size()) { PII v=E[u][i]; if (!v.second) continue; --dee[v.first]; if (!dee[v.first]) Q.push(v.first), mark[v.first]=true; } } } int main () { int T, x, y, res=0, tmp, sum=0; scanf("%d%d",&n,&m); s=n*m; t=n*m+1; FO(i,0,n) FO(j,0,m) { scanf("%d%d",&cost[i*m+j],&T); while (T--) { scanf("%d%d",&x,&y); E[x*m+y].pb(mp(i*m+j,0)); E[i*m+j].pb(mp(x*m+y,1)); ++dee[x*m+y]; } } FO(i,0,n) FO(j,0,m-1) FO(k,j+1,m) E[i*m+j].pb(mp(i*m+k,0)), E[i*m+k].pb(mp(i*m+j,1)), ++dee[i*m+j]; Topsort(); FO(i,0,n*m) { if (!mark[i]) continue; if (cost[i]>=0) res+=cost[i], add_edge(s,i,cost[i]); else add_edge(i,t,-cost[i]); FO(j,0,E[i].size()) { PII v=E[i][j]; if (v.second||!mark[v.first]) continue; add_edge(i,v.first,INF); } } while (bfs()) while (tmp=dfs(s,INF)) sum+=tmp; printf("%d\n",res-sum); return 0; }
BZOJ 1565 植物大战僵尸(拓扑排序+最大权闭合子图)
标签:最大 class namespace void algorithm cond 拓扑 main putc
原文地址:http://www.cnblogs.com/lishiyao/p/6785353.html