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-

# define MOD

# define INF

# 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<<,l,mid

# define rch p<<|,mid+,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=, flag=;

    char ch;

    if((ch=getchar())=='-') flag=;

    else if(ch>=''&&ch<='') res=ch-'';

    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');

    return flag?-res:res;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

struct Node{int x, y;}node[];

struct Edge{int to, next, cap, flow, cost;}edge[];

int head[N], tol, pre[N], dis[N], nn;

bool vis[N];

queue<int>q;

void init(int n){nn=n; tol=; mem(head,-);}

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=; edge[tol].next=head[u]; head[u]=tol++;

    edge[tol].to=u; edge[tol].cap=; edge[tol].cost=-cost; edge[tol].flow=; edge[tol].next=head[v]; head[v]=tol++;

}

bool spfa(int s, int t){

    FO(i,,nn) dis[i]=INF, vis[i]=false, pre[i]=-;

    dis[s]=; vis[s]=true; q.push(s);

    while (!q.empty()) {

        int u=q.front(); q.pop(); vis[u]=false;

        for (int i=head[u]; i!=-; 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]==-) return false;

    else return true;

}

int minCostMaxflow(int s, int t, int &cost){

    int flow=; cost=;

    while (spfa(s,t)) {

        int Min=INF;

        for (int i=pre[t]; i!=-; i=pre[edge[i^].to]) {

            if (Min>edge[i].cap-edge[i].flow) Min=edge[i].cap-edge[i].flow;

        }

        for (int i=pre[t]; i!=-; i=pre[edge[i^].to]) {

            edge[i].flow+=Min; edge[i^].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=; ss=*n+; t=*n+;

    init(t+);

    addedge(s,ss,,);

    FOR(i,,n) {

        scanf("%d%d",&node[i].x,&node[i].y);

        addedge(ss,i,,); addedge(i,n+i,,-); addedge(i,n+i,,); addedge(n+i,t,,);

    }

    sort(node+,node+n+,comp);

    FOR(i,,n) {

        int mi=INF, now=;

        FOR(j,i+,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,,);

        }

    }

    int ans;

    minCostMaxflow(s,t,ans);

    printf("%d\n",-ans);

    return ;

}

巴特西

BZOJ 1930 吃豆豆(费用流)

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