﻿ ZOJ3508 The War 贪心，最大流

### ZOJ3508 The War 贪心，最大流

A war had broken out because a sheep from your kingdom ate some grasses which belong to your neighboring kingdom. The counselor of your kingdom had to get prepared for this war. There are N (1 <= N <= 2500) unarmed soldier in your kingdom and there are M (1 <= M <= 40000) weapons in your arsenal. Each weapon has a weight W (1 <= W <= 1000), and for soldier i, he can only arm the weapon whose weight is between minWi and maxWi ( 1 <= minWi <= maxWi <= 1000). More armed soldier means higher success rate of this war, so the counselor wants to know the maximal armed soldier he can get, can you help him to win this war?

Input

There multiple test cases. The first line of each case are two integers N, M. Then the following N lines, each line contain two integers minWi, maxWi for each soldier. Next M lines, each line contain one integer W represents the weight of each weapon.

Output
```3 3
1 5
3 7
5 10
4
8
9
2 2
5 10
10 20
4
21```
Sample Output
```2
0```

#### 常规解法是贪心，但是在复习最大流的写法，因此用ｓａｐ来写的。思路是很好想的

```#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<memory.h>
#include<cmath>
using namespace std;
const int M=2510000;
int s,t,cnt;;
int a[4000],b[4000],num[1010];;
const int inf=1000000;
int vd[4001],dis[4001],ans,m,n;
int Laxt[M],Next[M],Val[M],To[M];
void _update()
{
memset(vd,0,sizeof(vd));
memset(dis,0,sizeof(dis));
memset(num,0,sizeof(num));
memset(Laxt,0,sizeof(Laxt));
ans=0; cnt=2;
s=0; t=1000+n+1;
}
{
Next[cnt]=Laxt[u];
Laxt[u]=cnt;
To[cnt]=v;
Val[cnt++]=w;

Next[cnt]=Laxt[v];
Laxt[v]=cnt;
To[cnt]=u;
Val[cnt++]=0;
}
int dfs(int u,int flow)
{
int temp,delta;
if(u==t)return flow;
delta=0;
for(int i=Laxt[u];i>0;i=Next[i])
if(Val[i]>0 && dis[u]==dis[To[i]]+1)
{
temp=dfs(To[i],min(flow-delta,Val[i]));
Val[i]-=temp;
Val[i^1]+=temp;

delta+=temp;
if(delta==flow||dis[s]>t) return delta;
}
vd[dis[u]]--;
if(vd[dis[u]]==0)dis[s]=t+1;
dis[u]++;
vd[dis[u]]++;
return delta;
}
void _work()
{
int i,j,x,y;
for(i=1;i<=n;i++) scanf("%d%d",&a[i],&b[i]);
for(i=1;i<=m;i++) {
scanf("%d",&x);
num[x]++;
}
for(i=1;i<=1000;i++)
if(num[i]>0)
for(i=1;i<=1000;i++)
for(j=1;j<=n;j++){
if(i>=a[j]&&i<=b[j])
}
for(j=1;j<=n;j++)
int temp=0;
while(dis[s]<t+1)
{
int flow=dfs(s,inf);
ans+=flow;
}
printf("%d
",ans);
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
_update();
_work();
}
return 0;
}```