链接：

https://vjudge.net/problem/HDU-3572

题意：

Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory has to start processing it at or after day Si, process it for Pi days, and finish the task before or at day Ei. A machine can only work on one task at a time, and each task can be processed by at most one machine at a time. However, a task can be interrupted and processed on different machines on different days.
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.

思路：

特殊建图， 源点到每个人连线，每个人到每一天连线，保证每个人每天只能用一台机器。
再求最大流，比较最大流和所需要的流即可。

代码： 
#include <iostream>
#include <queue>
#include <vector>
#include <memory.h>
using namespace std;
typedef long long LL;

const int MAXN = 5e2+10;
const int INF = 1<<30;

struct Edge
{
    int from, to, cap;
    Edge(int f, int t, int c):from(f), to(t), cap(c){}
};
vector<Edge> edges;
vector<int> G[MAXN*4];
int Dis[MAXN*4];
int Vis[MAXN];
int n, m;
int s, t;

bool Bfs()
{
    //Bfs构造分层网络
    memset(Dis, -1, sizeof(Dis));
    queue<int> que;
    que.push(s);
    Dis[s] = 0;
    while (!que.empty())
    {
        int u = que.front();
        que.pop();
        for (int i = 0;i < G[u].size();i++)
        {
            Edge & e = edges[G[u][i]];
            if (e.cap > 0 && Dis[e.to] == -1)
            {
                que.push(e.to);
                Dis[e.to] = Dis[u]+1;
            }
        }
    }
    return (Dis[t] != -1);
}

int Dfs(int u, int flow)
{
    //flow 表示当前流量上限
    if (u == t)
        return flow;
    int res = 0;
    for (int i = 0;i < G[u].size();i++)
    {
        Edge & e = edges[G[u][i]];
        if (e.cap > 0 && Dis[u]+1 == Dis[e.to])
        {
            int tmp = Dfs(e.to, min(flow, e.cap)); //  递归计算顶点 v
            flow -= tmp;
            e.cap -= tmp;
            res += tmp;
            edges[G[u][i]^1].cap += tmp;
            if (flow == 0)
                break;
        }
    }
    if (res == 0)
        Dis[u] = -1;
    return res;
}

int MaxFlow()
{
    int res = 0;
    while (Bfs())
    {
        res += Dfs(0, INF);
    }
    return res;
}

void Insert(int l, int r, int c)
{
    // 正反向插边
    edges.emplace_back(l, r, c);
    edges.emplace_back(r, l, 0);
    G[l].push_back(edges.size()-2);
    G[r].push_back(edges.size()-1);
}

void Init()
{
    edges.clear();
    for (int i = 0;i <= t;i++)
        G[i].clear();
}

int main()
{
    int l, r, c;
    int times, cnt = 0;
    scanf("%d", &times);
    while (times--)
    {
        scanf("%d%d", &n, &m);
        s = 0, t = 500+n+1;
        Init();
        int sum = 0;
        for (int i = 1;i <= n;i++)
        {
            scanf("%d%d%d", &c, &l, &r);
            Insert(0, i, c);
            for (int j = l;j <= r;j++)
            {
                Insert(i, n+j, 1);
                Vis[j] = 1;
            }
            sum += c;
        }
        for (int i = 1;i <= 500;i++)
            if (Vis[i])
                Insert(n+i, t, m);
        int res = MaxFlow();
        if (res == sum)
            printf("Case %d: Yes\n\n", ++cnt);
        else
            printf("Case %d: No\n\n", ++cnt);
    }

    return 0;
}