JZOJ 5329. 【NOIP2017提高A组模拟8.22】时间机器

5329. 【NOIP2017提高A组模拟8.22】时间机器

(File IO): input:machine.in output:machine.out

Time Limits: 2000 ms Memory Limits: 262144 KB

Description

Input

Output

Sample Input

3

2 2

1 4 2

3 5 1

1 4 2

2 5 1

3 2

1 3 1

2 4 1

3 5 1

1 3 2

2 5 1

2 2

1 2 2

1 2 1

1 2 1

1 2 2

Sample Output

Yes

No

Yes

Data Constraint

Hint

题解

贪心

将电阻和节点按左端点从小到大排序

按顺序考虑每一种节点

每次贪心选左端点在节点左边，右端点尽量靠近节点的电阻

用set或map维护一下左端点符合条件的右端点即可

代码

#include<cstdio>
#include<algorithm>
#include<map>
#define N 100010
using namespace std;
struct point{
    long x,y,z,type;
}a[N];
map<long,long>b;
bool cmp(point a,point b)
{
    if(a.x!=b.x)
        return a.x<b.x;
    else
        return a.type>b.type;
}
int main()
{   long tot,n,m,i;
    bool t;
    freopen("machine.in","r",stdin);
    freopen("machine.out","w",stdout);
    scanf("%ld",&tot);
    while(tot--){
        scanf("%ld%ld",&n,&m);
        for(i=1;i<=n;i++){
            scanf("%ld%ld%ld",&a[i].x,&a[i].y,&a[i].z);
            a[i].type=0;
        }
        for(i=1;i<=m;i++){
            scanf("%ld%ld%ld",&a[i+n].x,&a[i+n].y,&a[i+n].z);
            a[i+n].type=1;
        }
        sort(a+1,a+n+m+1,cmp);
        b.clear();
        t=true;
        for(i=1;i<=n+m;i++)
            if(a[i].type){
                if(!b[a[i].y])
                    b[a[i].y]=a[i].z;
                else
                    b[a[i].y]+=a[i].z;
            }else{
                while(a[i].z){
                    map<long,long>::iterator iter=b.lower_bound(a[i].y);
                    if(iter==b.end()){
                        t=false;
                        break;
                    }
                    if(a[i].z<iter->second){
                        iter->second-=a[i].z;
                        a[i].z=0;
                    }else{
                        a[i].z-=iter->second;
                        b.erase(iter);
                    }
                }
                if(!t)break;
            }
        if(t)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}