【JZOJ5329】【NOIP2017提高A组模拟8.22】时间机器

Description

Data Constraint

Solution

这道题看着网络流能过，实际上直接贪心就好。我们将电阻和节点分别按左端点排序，枚举节点，维护一个指针将电阻按右端点加入set，对于一个节点，我们肯定要找右端点不小于节点右端点的且尽量靠左的电阻来满足。按照这个贪心思路来搞即可。

Code

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<set>
#define ll long long
using namespace std;
const ll maxn=1e7+5,maxn1=1e5+5,mo=1e9+7,mo1=1e15;
struct code{
int l,r,x;
}a[maxn1],b[maxn1];
struct code1{
int x,id;
bool friend operator < (code1 x,code1 y){
return x.x<y.x || x.x==y.x && x.id<y.id;
}
}g;
int n,m,i,t,j,k,l,x,y,z,num,test,s,bz;
ll ans;
set<code1>f;
bool cmp(code x,code y){
return x.l<y.l;
}
int main(){
freopen("machine.in","r",stdin);freopen("machine.out","w",stdout);
scanf("%d",&test);
while (test--){
scanf("%d%d",&n,&m);s=n+m+1;f.clear();
for (i=1;i<=n;i++)scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].x);
for (i=1;i<=m;i++)scanf("%d%d%d",&b[i].l,&b[i].r,&b[i].x);
sort(a+1,a+n+1,cmp);sort(b+1,b+m+1,cmp);
j=1;g.x=mo;f.insert(g);bz=0;
for (i=1;i<=n;i++){
while (b[j].l<=a[i].l && j<=m) g.x=b[j].r,g.id=j++,f.insert(g);
g.x=a[i].r;g.id=0;
while (a[i].x){
g=*f.upper_bound(g);
if (g.x==mo) break;
t=min(b[g.id].x,a[i].x);
b[g.id].x-=t;a[i].x-=t;
if (!b[g.id].x) f.erase(g);
}
if (a[i].x){
bz=1;break;
}
}
if (!bz) printf("Yes\n");
else printf("No\n");
}
}