hdu 4274 Spy's Work(水题)
Spy's Work
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1266 Accepted Submission(s): 388
To get some information about ICPC, I have learned a lot about it. ICPC has N staffs now (numbered from 1 to N, and boss is 1), and anybody has at most one superior. To increase the efficiency of the whole company, the company contains N departments and the
ith department is led by the ith staff. All subordinates of the ith staff are also belong to the ith department.
Last week, we hire a spy stealing into ICPC to get some information about salaries of staffs. Not getting the detail about each one, the spy only gets some information about some departments: the sum of the salaries of staff s working for the ith department
is less than (more than or equal to) w. Although the some inaccurate information, we can also get some important intelligence from it.
Now I only concerned about whether the spy is telling a lie to us, that is to say, there will be some conflicts in the information. So I invite you, the talented programmer, to help me check the correction of the information. Pay attention, my dear friend,
each staff of ICPC will always get a salary even if it just 1 dollar!
The first line is an integer N. (1 <= N <= 10,000)
Each line i from 2 to N lines contains an integer x indicating the xth staff is the ith staff's superior(x<i).
The next line contains an integer M indicating the number of information from spy. (1 <= M <= 10,000)
The next M lines have the form like (x < (> or =) w), indicating the sum of the xth department is less than(more than or equal to) w (1 <= w <=100,000,000)
5
1
1
3
3
3
1 < 6
3 = 4
2 = 2 5
1
1
3
3
3
1 > 5
3 = 4
2 = 2
Lie
True
pid=4273" target="_blank">4273
pid=4270" target="_blank">4270
4269然后告诉你某棵子树权值和的范围。然后问你有没有矛盾。
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int maxn=10010;
typedef long long ll;
const ll INF=1e14;
int fa[maxn];
ll ns[maxn],nb[maxn],sz[maxn],ss[maxn],sb[maxn];
int main()
{
int n,i,x,m,w,flag;
char op[10]; while(~scanf("%d",&n))
{
fa[1]=flag=0;
for(i=1;i<=n;i++)
{
sz[i]=1;
ns[i]=-INF,nb[i]=INF;
ss[i]=sb[i]=0;
}
for(i=2;i<=n;i++)
{
scanf("%d",&x);
fa[i]=x;
}
scanf("%d",&m);
for(i=1;i<=m;i++)
{
scanf("%d%s%d",&x,op,&w);
if(op[0]=='=')
nb[x]=ns[x]=w;
else if(op[0]=='<')
nb[x]=min(nb[x],(ll)w-1);
else
ns[x]=max(ns[x],(ll)w+1);
}
for(i=n;i>=1;i--)
{
if(flag)
break;
ns[i]=max(ns[i],sz[i]);
if(ns[i]>nb[i]||nb[i]<=ss[i])
flag=1;
ns[i]=max(ns[i],ss[i]+1);
if(ns[i]>nb[i])
flag=1;
sz[fa[i]]+=sz[i];
ss[fa[i]]+=ns[i];
sb[fa[i]]+=nb[i];
}
if(flag)
printf("Lie\n");
else
printf("True\n");
}
return 0;
}
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