标签：
网络流，Dinic G++ 964ms水过，SAP 620ms AC
源点到每一个防晒霜建边，容量为cover[i]；
每一个防晒霜与它能保护的奶牛之间连一条边，容量为1；
每一个奶牛连到汇点一条边，容量为1。
最大流就是答案。
Dinic模板：
#include<cstdio> #include<cstring> #include<string> #include<cmath> #include<vector> #include<queue> #include<algorithm> using namespace std; const int maxn = 6000 + 10; const int INF = 0x7FFFFFFF; struct Edge { int from, to, cap, flow; Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f) {} }; vector<Edge>edges; vector<int>G[maxn]; bool vis[maxn]; int d[maxn]; int cur[maxn]; int n, m, s, t; void init() { for (int i = 0; i < maxn; i++) G[i].clear(); edges.clear(); } void AddEdge(int from, int to, int cap) { edges.push_back(Edge(from, to, cap, 0)); edges.push_back(Edge(to, from, 0, 0)); int w = edges.size(); G[from].push_back(w - 2); G[to].push_back(w - 1); } bool BFS() { memset(vis, 0, sizeof(vis)); queue<int>Q; Q.push(s); d[s] = 0; vis[s] = 1; while (!Q.empty()) { int x = Q.front(); Q.pop(); for (int i = 0; i<G[x].size(); i++) { Edge e = edges[G[x][i]]; if (!vis[e.to] && e.cap>e.flow) { vis[e.to] = 1; d[e.to] = d[x] + 1; Q.push(e.to); } } } return vis[t]; } int DFS(int x, int a) { if (x == t || a == 0) return a; int flow = 0, f; for (int &i = cur[x]; i<G[x].size(); i++) { Edge e = edges[G[x][i]]; if (d[x]+1 == d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0) { edges[G[x][i]].flow+=f; edges[G[x][i] ^ 1].flow-=f; flow+=f; a-=f; if(a==0) break; } } if(!flow) d[x] = -1; return flow; } int dinic(int s, int t) { int flow = 0; while (BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow; } int C,L; int minSPF[maxn],maxSPF[maxn]; int SPF[maxn]; bool judge(int a,int b) { if(SPF[a]>=minSPF[b]&&SPF[a]<=maxSPF[b]) return 1; return 0; } int main() { //0 源点 C+L+1汇点 1--L 防晒霜，L+1--L+C奶牛 scanf("%d%d",&C,&L); init(); s=0;t=C+L+1; for(int i=1;i<=C;i++) scanf("%d%d",&minSPF[i],&maxSPF[i]); for(int i=1;i<=L;i++) { int x; scanf("%d%d",&SPF[i],&x); AddEdge(0,i,x); } for(int i=1;i<=L;i++) for(int j=1;j<=C;j++) if(judge(i,j)) AddEdge(i,j+L,1); for(int i=L+1;i<=C+L;i++) AddEdge(i,t,1); printf("%d\n",dinic(s,t)); return 0; }
SAP模板：
#include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> using namespace std; const int maxn=500000,maxm=3000000,INF=1<<30; int tot=0,S,T,n,m,cur[maxn],lay[maxn],gap[maxn],pre[maxn],V[maxm],Q[maxn],G[maxm],N[maxm],B[maxm],F[maxn]; bool dw(int &x,int y) { if (y< x) { x=y; return 1; } return 0; } void add(int a,int b,int c) { ++tot;V[tot]=b;G[tot]=c;N[tot]=F[a];F[a]=tot; ++tot;V[tot]=a;G[tot]=0;N[tot]=F[b];F[b]=tot; B[tot]=tot-1;B[tot-1]=tot; } void bfs(int n) { int u,hd=0,tl=1,p; memset(lay,127,sizeof(lay)); memset(gap,0,sizeof(gap)); Q[1]=T;lay[T]=0;gap[0]=1; while (hd!=tl) { u=Q[++hd]; for (p=F[u];p;p=N[p]) if (G[B[p]]>0&&lay[V[p]]>=n) { lay[V[p]]=lay[u]+1; gap[lay[V[p]]]++; Q[++tl]=V[p]; cur[V[p]]=B[p]; } } } inline int sap(int n) { int u=S,v,flow,ret=0,mL,cnt=0,x,y; memset(pre,0,sizeof(pre));bfs(n); u=S; while (lay[S]< n) { cnt++; for (int &p=cur[u];p;p=N[p]) if (G[p]>0&&lay[u]==lay[V[p]]+1) break; if (cur[u]) { v=V[cur[u]]; pre[v]=u; u=v; if (v==T) { flow=INF; x=T; while (x!=S) { x=pre[x]; dw(flow,G[cur[x]]); } ret+=flow; if (ret>=INF) return ret; x=T; while (x!=S) { x=pre[x]; G[cur[x]]-=flow; G[B[cur[x]]]+=flow; } u=S; } }else { mL=n; for (int p=F[u];p;p=N[p]) if (G[p]>0&&dw(mL,lay[V[p]]+1)) cur[u]=p; gap[lay[u]]--; if (gap[lay[u]]==0) return ret; gap[mL]++;lay[u]=mL; if (u!=S) u=pre[u]; } } return ret; } int C,L; int minSPF[maxn],maxSPF[maxn]; int SPF[maxn]; bool judge(int a,int b) { if(SPF[a]>=minSPF[b]&&SPF[a]<=maxSPF[b]) return 1; return 0; } int main() { //0 源点 C+L+1汇点 1--L 防晒霜，L+1--L+C奶牛 scanf("%d%d",&C,&L); S=0;T=C+L+1; for(int i=1;i<=C;i++) scanf("%d%d",&minSPF[i],&maxSPF[i]); for(int i=1;i<=L;i++) { int x; scanf("%d%d",&SPF[i],&x); add(0,i,x); } for(int i=1;i<=L;i++) for(int j=1;j<=C;j++) if(judge(i,j)) add(i,j+L,1); for(int i=L+1;i<=C+L;i++) add(i,T,1); printf("%d\n",sap(C+L+1)); return 0; }
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原文地址：http://www.cnblogs.com/zufezzt/p/5169151.html
