hdu 4274 Spy's Work（水题）

Posted 2020-06-11

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

Problem Description

I‘m a manager of a large trading company, called ACM, and responsible for the market research. Recently, another trading company, called ICPC, is set up suddenly. It‘s obvious that we are the competitor to each other now!
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!

 

Input

There are multiple test cases.
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)

 

Output

For each test case, output "True" if the information has no confliction; otherwise output "Lie".

 

Sample Input

5
1
1
3
3
3
1 < 6
3 = 4
2 = 2

5
1
1
3
3
3
1 > 5
3 = 4
2 = 2

 

Sample Output

Lie
True

 

Source

2012 ACM/ICPC Asia Regional Changchun Online

 

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 题意：

给你一棵树。

然后告诉你某棵子树权值和的范围。然后问你有没有矛盾。

思路：

每一个结点维护两个值ns[i],nb[i]存当前子树的权值范围ns[i]<=val<=nb[i]从儿子到父亲一层一层更新父亲的范围。中间推断下是否冲突。

具体见代码：

#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;
}