【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 longusing 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");    }}