zoj 3500 The War

The War

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 areN (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

For each case, output one integer represents the maximal number of armed soldier you can get.

Sample Input

3 3
1 5
3 7
5 10
4
8
9
2 2
5 10
10 20
4
21

Sample Output

2
0

分析：武器有40000件，但只有1000种，可以压缩后建图

代码：
#include<cstdio>
#include<cstring>
const int N=4000;
const int M=2500000;
const int INF=0x7fffffff;
int tol,ans,s,t;
int head[N],arc[N],pre[N],dis[N],gap[N];
struct node
{
	int y,f,nxt;
}edge[M];
void add(int x,int y,int f)
{
	edge[tol].y=y;
	edge[tol].f=f;
	edge[tol].nxt=head[x];
	head[x]=tol++;
	edge[tol].y=x;
	edge[tol].f=0;
	edge[tol].nxt=head[y];
	head[y]=tol++;
}
void bfs(int n)
{
	int Q[N],front=0,rear=0,i,u,v;
	bool vis[N];
	memcpy(arc,head,sizeof(head));
	memset(dis,0,sizeof(dis));
	memset(vis,0,sizeof(vis));
	memset(gap,0,sizeof(gap));
	for(i=0;i<n;i++)dis[i]=n;
	vis[t]=1,dis[t]=0,Q[rear++]=t;
	while(front<rear)
	{
		u=Q[front++];
		for(i=head[u];i!=-1;i=edge[i].nxt)
		{
			v=edge[i].y;
			if(!vis[v]&&edge[i^1].f)
			{
				vis[v]=1;
				Q[rear++]=v;
				dis[v]=dis[u]+1;
				gap[dis[v]]++;
			}
		}
	}
}
void sap(int n)
{
	bfs(n);
	int arg=INF,u=pre[s]=s;
	while(dis[s]<n)
	{
L:
		for(int& i=arc[u];i!=-1;i=edge[i].nxt)
		{
			int v=edge[i].y;
			if(edge[i].f&&dis[u]==dis[v]+1)
			{
				if(arg>edge[i].f)arg=edge[i].f;
				pre[v]=u;u=v;
				if(v==t)
				{
					ans+=arg;
					for(u=pre[u];v!=s;v=u,u=pre[u])
					{
						edge[arc[u]].f-=arg;
						edge[arc[u]^1].f+=arg;
					}
					arg=INF;
				}
				goto L;
			}
		}
		int min=n;
		for(int j=head[u];j!=-1;j=edge[j].nxt)
		{
			int v=edge[j].y;
			if(edge[j].f&&min>dis[v])
			{
				arc[u]=j;
				min=dis[v];
			}
		}
		if(--gap[dis[u]]==0)break;
		dis[u]=min+1;
		gap[dis[u]]++;
		u=pre[u];
	}
}
struct Soldier{
	int minw,maxw;
}so[2502];
int cw[1001];
int main()
{
	int n,m,i,j,w,mw;
	while(~scanf("%d%d",&n,&m)){
		ans=0,tol=0;
		memset(head,-1,sizeof(head));
		for(i=1;i<=n;i++)scanf("%d%d",&so[i].minw,&so[i].maxw);
		memset(cw,0,sizeof(cw));mw=0;
		while(m--){scanf("%d",&w);cw[w]++;if(w>mw)mw=w;}
		s=0,t=n+mw+1;
		for(i=1;i<=n;i++){
			add(s,i,1);
			for(j=so[i].minw;j<=so[i].maxw;j++){
				if(cw[j])add(i,n+j,INF);
			}
		}
		for(i=1;i<=mw;i++)if(cw[i])add(n+i,t,cw[i]);
		n=t+1;
		sap(n);
		printf("%d\n",ans);
	}
	return 0;
}
分析：另外此题贪心可解

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int w[1001];
struct node{
	int minw,maxw;
}a[2500];
bool cmp(node a,node b){
	if(a.maxw==b.maxw)return a.minw>b.minw;
	return a.maxw<b.maxw;
}
int main(){
	int n,m,i,j,x,f,ans;
	while(~scanf("%d%d",&n,&m)){
		for(i=0;i<n;i++)scanf("%d%d",&a[i].minw,&a[i].maxw);
		sort(a,a+n,cmp);
		memset(w,0,sizeof(w));
		for(i=0;i<m;i++)scanf("%d",&x),w[x]++;
		ans=0;
		for(i=0;i<n;i++){
			f=0;
			for(j=a[i].minw;j<=a[i].maxw;j++)if(w[j]){w[j]--;f=1;break;}
			ans+=f;
		}
		printf("%d\n",ans);
	}
	return 0;
}