poj 3614(网络流)

Sunscreen

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6672		Accepted: 2348

Description

To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they're at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPF_i ≤ 1,000; minSPF_i ≤ maxSPF_i ≤ 1,000) that will work. If the SPF rating is too low, the cow suffers sunburn; if the SPF rating is too high, the cow doesn't tan at all........

The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion, each bottle i with an SPF rating SPF_i (1 ≤ SPF_i ≤ 1,000). Lotion bottle i can cover cover_i cows with lotion. A cow may lotion from only one bottle.

What is the maximum number of cows that can protect themselves while tanning given the available lotions?

Input

* Line 1: Two space-separated integers: C and L
* Lines 2..C+1: Line i describes cow i's lotion requires with two integers: minSPF_i and maxSPF_i
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPF_i and cover_i

Output

A single line with an integer that is the maximum number of cows that can be protected while tanning

Sample Input

3 2
3 10
2 5
1 5
6 2
4 1

Sample Output

2

Source

题意：N头牛,第I头需要一个SPF的范围是MinSPF~MaxSPF,m个bottle,每个bottle能给C头牛提供定值为P的SPF,求最多有多少头牛可以得到合适的SPF.

题解:设立超级源点S和超级汇点T，S向每个Bottle引一条容量为P的边，如果牛可以忍受这个bottle的SPF，那么就连一条容量为1的边，最后所有的牛像汇点引一条容量为1的边，求最大流即可。

#include <stdio.h>
#include <algorithm>
#include <queue>
#include <string.h>
#include <math.h>
#include <iostream>
#include <math.h>
using namespace std;
const int N = ;
const int INF = ;
struct Edge
{
    int v,next;
    int w;
} edge[N*N];
struct Cow{
    int minSPF,maxSPF;
}cow[N/];
int head[N];
int level[N];
int tot;
void init()
{
    memset(head,-,sizeof(head));
    tot=;
}
void addEdge(int u,int v,int w,int &k)
{
    edge[k].v = v,edge[k].w=w,edge[k].next=head[u],head[u]=k++;
    edge[k].v = u,edge[k].w=,edge[k].next=head[v],head[v]=k++;
}
int BFS(int src,int des)
{
    queue<int >q;
    memset(level,,sizeof(level));
    level[src]=;
    q.push(src);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        if(u==des) return ;
        for(int k = head[u]; k!=-; k=edge[k].next)
        {
            int v = edge[k].v;
            int w = edge[k].w;
            if(level[v]==&&w!=)
            {
                level[v]=level[u]+;
                q.push(v);
            }
        }
    }
    return -;
}
int dfs(int u,int des,int increaseRoad){
    if(u==des||increaseRoad==) return increaseRoad;
    int ret=;
    for(int k=head[u];k!=-;k=edge[k].next){
        int v = edge[k].v,w=edge[k].w;
        if(level[v]==level[u]+&&w!=){
            int MIN = min(increaseRoad-ret,w);
            w = dfs(v,des,MIN);
            if(w > )
            {
                edge[k].w -=w;
                edge[k^].w+=w;
                ret+=w;
                if(ret==increaseRoad) return ret;
            }
            else level[v] = -;
            if(increaseRoad==) break;
        }
    }
    if(ret==) level[u]=-;
    return ret;
}
int Dinic(int src,int des)
{
    int ans = ;
    while(BFS(src,des)!=-) ans+=dfs(src,des,INF);
    return ans;
}
int main()
{
    int c,l;
    scanf("%d%d",&c,&l);
    init();
    int src = ,des = c+l+;
    for(int i=; i<=c; i++)
    {
        scanf("%d%d",&cow[i].minSPF,&cow[i].maxSPF);
        addEdge(i+l,des,,tot);
    }
    for(int i=; i<=l; i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        addEdge(src,i,b,tot);
        for(int j=; j<=c; j++)
        {
            if(a>=cow[j].minSPF&&a<=cow[j].maxSPF)
            {
                addEdge(i,j+l,,tot);
            }
        }
    }
    printf("%d\n",Dinic(src,des));
}