hdu 3572

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3572

题目思路：裸的网络流，只要建图小心点就行了，建图能少边就少边，能减少复杂度，另外，数组一定要开大，玄学TL

AC代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;

#define maxn 1000
#define maxm 500000
#define inf 0x3f3f3f3f
int first[maxn];
int edge_num;
int lev[maxn];
int vis[maxn];
int mark[maxn];
struct edge{
    int u,v;
    int cap;
    int next;
}e[maxm];
struct node{
    int start;
    int end;
    int process;
    node(int p=0,int s=0,int e=0)
    {
        process=p;
        start=s;
        end=e;
    }
};
vector<node>vec;

void init()
{
    edge_num=0;
    memset(first,-1,sizeof(first));
}

void add_edge(int u,int v,int c)
{
    e[edge_num].u=u;
    e[edge_num].v=v;
    e[edge_num].cap=c;
    e[edge_num].next=first[u];
    first[u]=edge_num++;
    e[edge_num].u=v;
    e[edge_num].v=u;
    e[edge_num].cap=0;
    e[edge_num].next=first[v];
    first[v]=edge_num++;
}

int Dinic_spath(int s,int t)
{
    queue<int>q;
    q.push(s);
    memset(lev,-1,sizeof(lev));
    memset(vis,0,sizeof(vis));
    lev[s]=0;
    while(!q.empty())
    {
        int u=q.front();
        vis[u]=0;
        q.pop();
        for(int k=first[u];k!=-1;k=e[k].next)
        {
            int v=e[k].v;
            if(lev[v]==-1&&e[k].cap>0)
            {
                lev[v]=lev[u]+1;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    return lev[t]!=-1;
}

int Dinic_dfs(int u,int t,int flow)
{
    int cost=0;
    if(u==t) return flow;
    for(int k=first[u];k!=-1;k=e[k].next)
    {
        int v=e[k].v;
        if(lev[v]==lev[u]+1&&e[k].cap>0)
        {
            int minn=Dinic_dfs(v,t,min(e[k].cap,flow-cost));
            if(minn>0)
            {
                e[k].cap=e[k].cap-minn;
                e[k^1].cap=e[k^1].cap+minn;
                cost+=minn;
                if(cost==flow) break;
            }
            else
                lev[v]=-1;
        }
    }
    return cost;
}

int main()
{
    int t;
    scanf("%d",&t);
    int ca=1;
    while(t--)
    {
        init();
        int n,m;
        scanf("%d%d",&n,&m);
        int sum=0;
        int src=0;
        int dst=500+n+1;
        memset(mark,0,sizeof(mark));
        for(int i=1;i<=n;++i)
        {
            int p,s,e;
            scanf("%d%d%d",&p,&s,&e);
            add_edge(500+i,dst,p);
            sum+=p;
            for(int j=s;j<=e;++j)
            {
                mark[j]=1;
                add_edge(j,500+i,1);
            }
        }
        for(int i=1;i<=500;++i)
            if(mark[i]==1)
                add_edge(src,i,m);
        int f=0;
        while(Dinic_spath(0,500+n+1))
        {
            f+=Dinic_dfs(0,500+n+1,inf);
        }
        if(f==sum)
            printf("Case %d: Yes\n\n",ca++);
        else
            printf("Case %d: No\n\n",ca++);
    }
    return 0;

}