BZOJ 1930 吃豆豆(费用流)

首先这题的两条线不相交的限制可以去掉，因为如果相交的话把点换一换是不影响最终结果的。

剩下的费用流建图是显然的，把点拆为两个，建立超级源点s和源点ss汇点t，连边(s,ss,2,0). 对于每个点，连边(ss,i,1,0), (i,i',1,1),(i',t,1,0).

这样跑一遍费用流就行了，然而此题的边数可以达到n^2.无疑是OLE的。需要优化。

容易发现，对于点(i,j),(j,k),(i,k).如果这些点都可以互相到达的话，那么(i,k)这条边是不必要的。因为通过j到达k是不会比结果劣的。

所以启发我们将点按x坐标以第一关键字排序，y第二关键字排序。对于相同的列，选择大于前一列的y坐标且最小的点连边。

这样建边之后还存在一个问题，可能有个点两个吃豆人都需要经过，于是再建边(i,i',1,0),(i',j,2,0).

# 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 3.1415926535
# define eps 1e-9
# define MOD 1000000007
# 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 res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=4005;
//Code begin...

struct Node{int x, y;}node[2005];
struct Edge{int to, next, cap, flow, cost;}edge[1000005];
int head[N], tol, pre[N], dis[N], nn;
bool vis[N];
queue<int>q;

void init(int n){nn=n; tol=0; mem(head,-1);}
void addedge(int u, int v, int cap, int cost){
    edge[tol].to=v; edge[tol].cap=cap; edge[tol].cost=cost; edge[tol].flow=0; edge[tol].next=head[u]; head[u]=tol++;
    edge[tol].to=u; edge[tol].cap=0; edge[tol].cost=-cost; edge[tol].flow=0; edge[tol].next=head[v]; head[v]=tol++;
}
bool spfa(int s, int t){
    FO(i,0,nn) dis[i]=INF, vis[i]=false, pre[i]=-1;
    dis[s]=0; vis[s]=true; q.push(s);
    while (!q.empty()) {
        int u=q.front(); q.pop(); vis[u]=false;
        for (int i=head[u]; i!=-1; i=edge[i].next) {
            int v=edge[i].to;
            if (edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost) {
                dis[v]=dis[u]+edge[i].cost; pre[v]=i;
                if (!vis[v]) vis[v]=true, q.push(v);
            }
        }
    }
    if (pre[t]==-1) return false;
    else return true;
}
int minCostMaxflow(int s, int t, int &cost){
    int flow=0; cost=0;
    while (spfa(s,t)) {
        int Min=INF;
        for (int i=pre[t]; i!=-1; i=pre[edge[i^1].to]) {
            if (Min>edge[i].cap-edge[i].flow) Min=edge[i].cap-edge[i].flow;
        }
        for (int i=pre[t]; i!=-1; i=pre[edge[i^1].to]) {
            edge[i].flow+=Min; edge[i^1].flow-=Min; cost+=edge[i].cost*Min;
        }
        flow+=Min;
    }
    return flow;
}
bool comp(Node a, Node b){
    if (a.x==b.x) return a.y<b.y;
    else return a.x<b.x;
}
int main ()
{
    int n, s, ss, t;
    scanf("%d",&n);
    s=0; ss=2*n+1; t=2*n+2;
    init(t+1);
    addedge(s,ss,2,0);
    FOR(i,1,n) {
        scanf("%d%d",&node[i].x,&node[i].y);
        addedge(ss,i,1,0); addedge(i,n+i,1,-1); addedge(i,n+i,1,0); addedge(n+i,t,1,0);
    }
    sort(node+1,node+n+1,comp);
    FOR(i,1,n) {
        int mi=INF, now=0;
        FOR(j,i+1,n) {
            if (now==node[j].x||node[j].y<node[i].y) continue;
            if (node[j].y<mi) mi=node[j].y, now=node[j].x, addedge(n+i,j,2,0);
        }
    }
    int ans;
    minCostMaxflow(s,t,ans);
    printf("%d
",-ans);
    return 0;
}