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 100010using 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;}