POJ 3614 Sunscreen

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;
}