The 2019 China Collegiate Programming Contest Harbin Site A - Artful Paintings 差分约束

#include<map>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
const int N=1e5+5;
const int M=1e5+5;
const int INF=0x3f3f3f3f;
int read()
{
	int res=0,ch,flag=0;
	if((ch=getchar())=='-')             //判断正负
		flag=1;
	else if(ch>='0'&&ch<='9')           //得到完整的数
		res=ch-'0';
	while((ch=getchar())>='0'&&ch<='9')
		res=res*10+ch-'0';
	return flag?-res:res;
}
int t;
int n,M1,M2;
int e[N],h[N],ne[N],w[N],idx;
int l1[N],r1[N],k1[N];
int l2[N],r2[N],k2[N];
inline void add(int u,int v,int c)
{
	e[idx]=v;
	w[idx]=c;
	ne[idx]=h[u];
	h[u]=idx++;
}
int dis[N];
bool vis[N];

inline bool spfa(int mid)
{
	queue<int> q;
	idx=0;
	memset(h,-1,sizeof h);
	memset(dis,-1,sizeof dis);
	memset(vis,0,sizeof vis);
	bool flag=0;
	for(int i=1; i<=n; i++)
	{
		add(i-1,i,0);
		add(i,i-1,-1);
	}
	//l r 不少于k1 
	//l-1  -> r  k1 
	for(int i=1; i<=M1; i++)
		add(l2[i]-1,r2[i],k2[i]);
	//l r 之外的 不少于k2
	//也就是 l r  之间 <=mid-k2
	//改成相同的就是
	//r -> l-1   mid-k2
	for(int i=1; i<=M2; i++)
		add(r1[i],l1[i]-1,k1[i]-mid);
	add(0,n,mid);
	add(n,0,-mid);
	dis[0]=0;
	q.push(0);
	vis[0]=1;
	while(!q.empty()&& !flag)
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(int i=h[u]; (~i)&&!flag; i=ne[i])
		{
			if(dis[e[i]]<dis[u]+w[i])
			{
				dis[e[i]]=dis[u]+w[i];
				if(dis[0]>0) 
					flag=1;
				if(!vis[e[i]])
				{
					q.push(e[i]);
					vis[e[i]]=1;
				}
			}
		}
	}
	return !flag;
}
int main()
{
	t=read();
	while(t--)
	{
		n=read(),M1=read(),M2=read();
		int l=0,r=n,mid,ans=n;
		for(int i=1; i<=M1; i++)
			l2[i]=read(),r2[i]=read(),k2[i]=read();
		for(int i=1; i<=M2; i++)
			l1[i]=read(),r1[i]=read(),k1[i]=read();
		while(l<r)
		{
			mid=(l+r)>>1;
			if(spfa(mid))
				r=mid;
			else
				l=mid+1;
		}
		cout<<r<<endl;
	}
	return 0;
}