HDU 3572——Task Schedule

最大流问题 

Dinic算法

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
const int maxint=1000001;
int n,m,s,t;

struct edgee
{
	int from,to,cap,flow;
};

vector<edgee> edges;
vector<int> G[1003];
bool vis[1003];
int dist[1003];
int cur[1003];

void add(int from,int to,int cap)
{
	edges.push_back((edgee){from,to,cap,0});
	edges.push_back((edgee){to,from,0,0});
	int k=edges.size();
	G[from].push_back(k-2);//偶数数为正向弧 
	G[to].push_back(k-1);
}

int max(int a,int b)
{
	return a>b?a:b;
}
int min(int a,int b)
{
	return a<b?a:b;
}

bool bfs()
{
	memset(vis,0,sizeof(vis));
	queue<int> q;
	dist[s]=0;
	vis[s]=1;
	q.push(s);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		int sz=G[u].size();
		for(int i=0;i<sz;i++)
		{
			edgee& e=edges[G[u][i]];
			if(!vis[e.to] && e.cap>e.flow)
			{
				vis[e.to]=1;
				dist[e.to]=dist[u]+1;
				q.push(e.to);
			}
		}
	}
	return vis[t];
}

int dfs(int u,int low)
{
	if(u==t || low==0)
		return low;
	int flow=0,sz=G[u].size(),d;//&引用的妙用
	for(int& i=cur[u];i<sz;i++) //正在考虑的最后一条边开始考虑（效率关键）
	{
		edgee& e=edges[G[u][i]];
		if(dist[u]+1==dist[e.to] && (d=dfs(e.to,min(e.cap-e.flow,low)))>0 )
		{
			e.flow+=d;
			edges[G[u][i]^1].flow-=d;
			flow+=d;
			low-=d;
			if(low==0)
				break;
		}
	}
	return flow;
}

int Dinic()
{
	int flow=0;
	while(bfs())
	{
		memset(cur,0,sizeof(cur));
		flow+=dfs(s,maxint);
	}
	return flow;
}

int main()
{
	int T;
	int i,j;
	int maxn;
	int flag=1;
	int sum;
	cin>>T;
	while(T--)
	{
		sum=0;
		int n,m;
		maxn=0;
		cin>>n>>m;
		edges.clear();
		for(i=0;i<=1000;i++)
			G[i].clear();		
		for(i=1;i<=n;i++)
		{
			int pi,ci,ei;
			cin>>pi>>ci>>ei;
			sum+=pi;
			maxn=max(maxn,ei);
			add(0,i,pi);
			for(j=ci;j<=ei;j++)
				add(i,n+j,1);	
		}
		for(i=1;i<=maxn;i++)
			add(n+i,maxn+n+1,m);
		s=0,t=maxn+n+1;
		int tmp=Dinic();
		if(sum==tmp)
    		printf("Case %d: Yes\n\n",flag++);
		else
			printf("Case %d: No\n\n",flag++);
	}
	return 0;
}