hdu 5406 CRB and Apple, 2015多校联合训练赛，费用流

CRB and Apple

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 336    Accepted Submission(s): 101

Problem Description In Codeland there are many apple trees.

One day CRB and his girlfriend decided to eat all apples of one tree.

Each apple on the tree has height and deliciousness.

They decided to gather all apples from top to bottom, so an apple can be gathered only when it has equal or less height than one just gathered before.

When an apple is gathered, they do one of the following actions.

1.  CRB eats the apple.

2.  His girlfriend eats the apple.

3.  Throw the apple away.

CRB(or his girlfriend) can eat the apple only when it has equal or greater deliciousness than one he(she) just ate before.

CRB wants to know the maximum total number of apples they can eat.

Can you help him?

Input There are multiple test cases. The first line of input contains an integer  T , indicating the number of test cases. For each test case:

The first line contains a single integer  N  denoting the number of apples in a tree.

Then  N  lines follow,  i -th of them contains two integers  Hi  and  Di  indicating the height and deliciousness of  i -th apple.

1 ≤  T  ≤ 48

1 ≤  N  ≤ 1000

1 ≤  Hi ,  Di  ≤  109

Output For each test case, output the maximum total number of apples they can eat.  

Sample Input

1
5
1 1
2 3
3 2
4 3
5 1
        

Sample Output

4
        

Author KUT（DPRK）  

Source 2015 Multi-University Training Contest 10

Accepted

5406

296MS

2204K

3031 B

G++

参考内容来自：

http://blog.csdn.net/mxymxy1994mxy/article/details/47818397

这篇博客中提供的思路大约边都为原来的50分之一了，优化好厉害。学习了。

http://blog.csdn.net/l_ecry/article/details/47830927

用栈优化时间快了4倍左右！！也是6666

对于spfa的优化，对于队列而言确实能够有不少的优化。如下：

http://www.cnblogs.com/cj695/archive/2012/07/27/2611215.html

对于这题建图：

建立s,t,  把i点拆为i,i+n,建立虚拟节点限制流量s1， 

s->s1  容量2，

s->i, i->t,容量1，

除了i->i+n的费用为-1，其他边费用都为0

对h从小到大排序，d从大到小排序，

边优化：对于a+n->x, a+n->y 如果x >= y 那么a+n不需要对y连边，因为通过x也可以到达y。

这就是博客1中单调块的思想。但是，会出现单调块的入口被占据的情况，

那么建边的同时建立一条a->x的路径，表示路过a点，直接访问a能到达的点。

费用流好久没用了，好神奇。！！！！

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define maxn 2017
#define maxm 200000
#define inf 10000000
using namespace std;
int head[maxn],tail;
int queue[maxn],pre[maxn],flag[maxn];
int dist[maxn],maxFlow[maxn];
struct Edge{
    int v,u,next,cost,w;
    Edge(){}//费用，权值
    Edge(int u,int v,int next,int cost,int w):u(u),v(v),next(next),cost(cost),w(w){}
} edge[maxm];
void add_edge(int u,int v,int cost,int w){
    edge[tail] = Edge(u,v,head[u],cost,w);
    head[u] = tail++;
    edge[tail] = Edge(v,u,head[v],-cost,0);
    head[v] = tail++;
}
void init(){
    memset(head,-1,sizeof(head));
    tail=0;
}
int SPFA(int start,int end){
    int i,u,v,front,rear;
    front = rear = 0;
    memset(flag,0,sizeof(flag));
    memset(dist,0x1f,sizeof(dist));
    memset(pre,-1,sizeof(pre));
    dist[start] = 0, dist[end] = inf ,flag[start]=1;
    maxFlow[start] = inf, queue[rear++] = start;
    while(front != rear){//增广
        u = queue[--rear];
        //if(front >= maxn) front = 0;
        flag[u] = 0;
        for(i = head[u]; i!=-1;i=edge[i].next){
            v=edge[i].v;
            if(edge[i].w&&dist[v]>dist[u]+edge[i].cost){
                dist[v]=dist[u]+edge[i].cost;
                maxFlow[v]=min(edge[i].w,maxFlow[u]);
                pre[v]=i;//记录边下标
                if(!flag[v]){
                    queue[rear++]=v;
                   // if(rear>=maxn)rear=0;
                    flag[v] =1;
                }
            }
        }
    }
    return dist[end] != inf;
}
//开始点，结束点
int MFMC(int start,int end){
    int min_cost = 0,v;
    while(SPFA(start,end)){
        v = end;
        while(pre[v]>=0){
            edge[pre[v]].w-=maxFlow[end];
            edge[pre[v]^1].w+=maxFlow[end];
            v=edge[pre[v]].u;
        }//跟新费用
        min_cost+=dist[end]*maxFlow[end];
    }
    return min_cost;
}
struct Node{
    int h,d;
};
Node p[1001];
int comp(Node a,Node b){
    if(a.h == b.h) return a.d > b.d;
    return a.h < b.h;
}
int main(){
    int t,n,h,d;
    //freopen("1001.in","r",stdin);
    //freopen("1001x.out","w",stdout);
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i = 0 ;i < n; i++){
            scanf("%d%d",&p[i].h,&p[i].d);
        }
        sort(p,p+n,comp);
        init();
        int s1 = n*2, S = s1+1, T = S+1;
        for(int i = 0;i < n; i++){
            add_edge(s1,i,0,2);
            add_edge(i+n,T,0,1);
            add_edge(i,i+n,-1,1);
        }
        add_edge(S,s1,0,2);
        for(int i = 0;i < n; i++){
            int x = -10;
            for(int j = i+1;j < n; j++){
                if(p[j].d <= p[i].d && p[j].d > x){
                    add_edge(i+n,j,0,1);
                    add_edge(i,j,0,1);
                    x = p[j].d;
                }
            }
        }
        int ans = 0;
        ans = MFMC(S,T);
        printf("%d\n",-ans);
    }
    return 0;
}