The topic link
uses "time particles" as the calculation result of the maximum flow.
Set the super source point to 0. The
customer point range is 1-204
time points 205-610.
Super sink point 615
Super source point is connected to all customers, and the capacity is the number of barbecues required * The time required for each piece of barbecue (that is, the total time particles that the customer needs to occupy) the
customer and the time point are connected, only when the time point is within the time period when the customer is waiting, the capacity is INF
each time point and meeting point Connection, capacity m
// #include <cstdio>
// #include <cstdlib>
// #include <cmath>
// #include <cstring>
// #include <iostream>
// #include <algorithm>
// #include <string>
// #include <vector>
// #include <queue>
// #include <stack>
// #include <maps>
// #include <list>
#include <bits/stdc++.h>
using namespace std;
/*
* 最大流 SAP 算法，用 GAP 优化后
* 先把流量限制赋值到 maps 数组
* 然后调用 SAP 函数求解
* 可选：导出路径
*/
#define MAXN 620
int maps[MAXN][MAXN]; // 存图
int pre[MAXN]; // 记录当前点的前驱
int level[MAXN]; // 记录距离标号
int gap[MAXN]; // gap常数优化
// vector<int> roads[MAXN]; // 导出的路径（逆序）
// int curRoads; // 导出的路径数
// 入口参数vs源点，vt汇点
int SAP(int vs, int vt)
{
memset(pre, -1, sizeof(pre));
memset(level, 0, sizeof(level));
memset(gap, 0, sizeof(gap));
gap[0] = vt;
int v, u = pre[vs] = vs, maxflow = 0, aug = INT_MAX;
// curRoads = 0;
while (level[vs] < vt)
{
// 寻找可行弧
for (v = 1; v <= vt; v++)
{
if (maps[u][v] > 0 && level[u] == level[v] + 1)
{
break;
}
}
if (v <= vt)
{
pre[v] = u;
u = v;
if (v == vt)
{
// int neck = 0; // Dnic 多路增广优化，下次增广时，从瓶颈边(后面)开始
aug = INT_MAX;
// 寻找当前找到的一条路径上的最大流 (瓶颈边)
for (int i = v; i != vs; i = pre[i])
{
// roads[curRoads].push_back(i); // 导出路径——可选
if (aug > maps[pre[i]][i])
{
aug = maps[pre[i]][i];
// neck = i; // Dnic 多路增广优化，下次增广时，从瓶颈边(后面)开始
}
}
// roads[curRoads++].push_back(vs); // 导出路径——可选
maxflow += aug;
// 更新残留网络
for (int i = v; i != vs; i = pre[i])
{
maps[pre[i]][i] -= aug;
maps[i][pre[i]] += aug;
}
// 从源点开始继续搜
u = vs;
// u = neck; // Dnic 多路增广优化，下次增广时，从瓶颈边(后面)开始
}
}
else
{
// 找不到可行弧
int minlevel = vt;
// 寻找与当前点相连接的点中最小的距离标号
for (v = 1; v <= vt; v++)
{
if (maps[u][v] > 0 && minlevel > level[v])
{
minlevel = level[v];
}
}
gap[level[u]]--; // (更新gap数组）当前标号的数目减1；
if (gap[level[u]] == 0)
break; // 出现断层
level[u] = minlevel + 1;
gap[level[u]]++;
u = pre[u];
}
}
return maxflow;
}
// 超级源点 0
// 顾客点 1 - 204
// 时间点 205 - 610
// 超级汇点 615
int n, m;
const int MaxPeople = 210;
const int Ed = 615;
struct costman
{
int b, e, need, time;
};
set<int> timelist;
costman costmanlist[MaxPeople];
void init()
{
memset(maps, 0, sizeof(maps));
set<int>::iterator iterl = timelist.begin();
set<int>::iterator iterr = timelist.begin();
iterr++;
int curiter = 0;
for (size_t i = 0; i < n; i++)
{
maps[0][i + 1] = costmanlist[i].need * costmanlist[i].time;
}
while (iterr != timelist.end())
{
maps[205 + curiter][Ed] = ((*iterr) - (*iterl)) * m;
for (size_t i = 0; i < n; i++)
{
if (costmanlist[i].b <= *iterl && costmanlist[i].e >= *iterr)
{
maps[i + 1][curiter + 205] = INT_MAX;
}
}
iterl++;
iterr++;
curiter++;
}
}
int main()
{
#ifdef ACM_LOCAL
freopen("./in.txt", "r", stdin);
freopen("./out.txt", "w", stdout);
#endif
ios::sync_with_stdio(false);
while (cin >> n >> m)
{
timelist.clear();
long long sum = 0;
for (size_t i = 0; i < n; i++)
{
cin >> costmanlist[i].b >> costmanlist[i].need >> costmanlist[i].e >> costmanlist[i].time;
sum += costmanlist[i].need * costmanlist[i].time;
timelist.insert(costmanlist[i].b);
timelist.insert(costmanlist[i].e);
}
init();
cout << (sum == SAP(0, Ed) ? "Yes" : "No") << endl;
}
return 0;
}