zoj 3500 The War

最新推荐文章于 2025-02-01 17:25:18 发布

原创最新推荐文章于 2025-02-01 17:25:18 发布 · 465 阅读

0 ·

CC 4.0 BY-SA版权

网络流专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一个关于战争资源分配的问题，具体涉及如何最大化武装士兵数量。通过两种不同的算法实现这一目标，一种是基于最大流的解决方案，另一种是采用贪心算法进行优化。

The War

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

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;
}