2019/9/2 school practice sessions examination report

A. God a graph theory problem (god.cpp)

Topic Link

The meaning of problems

Given an undirected graph, each point has the right to punctual, delete points is the cost of a weight not removed all nodes connected to it and , find the minimum cost to delete all points.

solution

Consider any two points connected \ (U \) and \ (V \) .

Set \ (U \) point with a weight of \ (A \) .

\ (V \) point with a weight of \ (B \) .

And \ (U \) weight value of the node and connected, by subtracting \ (V \) point weights, the result is \ (C \) .

And \ (V \) weight value of the node and connected, by subtracting \ (U \) point weights, the result is \ (D \) .

So, if you delete \ (U \) , then delete \ (v \) , the price for \ (b + c + d \) .

If you delete \ (v \) , then delete \ (U \) , a consideration of \ (A + c + d \) .

So if \ (A> b \) , should delete \ (U \) , then delete \ (v \) .

Otherwise, you should delete the \ (v \) , then delete \ (U \) .

That should delete \ (u, v \) greater weight in points.

Obviously this conclusion can be generalized to all points.

So greedy strategy is to delete large descending weights point.

Code

#include<bits/stdc++.h>
const int SIZE=200005;
int x[SIZE],head[SIZE],nex[SIZE],to[SIZE],Tot,Ans[SIZE];
void Link(int u,int v)
{
    nex[++Tot]=head[u];head[u]=Tot;to[Tot]=v;
    nex[++Tot]=head[v];head[v]=Tot;to[Tot]=u;
}

struct node
{
    int x,pos;
    bool operator <(const node &p)const
    {
        return x>p.x;
    }
}Tem[SIZE];

bool Removed[SIZE];
long long Tot_Ans;
int main()
{
    freopen("god.in","r",stdin);
    freopen("god.out","w",stdout);
    int n,m,u,v;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x[i]);
        Tem[i]=(node){x[i],i};
    }
    std::sort(Tem+1,Tem+1+n);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&u,&v);
        Link(u,v);
        Ans[u]+=x[v];
        Ans[v]+=x[u];
    }
    for(int o=1;o<=n;o++)
    {
        u=Tem[o].pos;
        for(int i=head[u];i;i=nex[i])
        {
            v=to[i];
            if(Removed[v])continue;
            Ans[v]-=x[u];
        }
        Tot_Ans+=Ans[u];
        Removed[u]=1;
    }
    printf("%lld",Tot_Ans);
    return 0;
}

B. Bitwise (bit.cpp)

Topic Link

The meaning of problems

Custom number \ (n \) value of \ (k \) is \ (| n | \) digits of the sum.

Seeking to meet less than \ (n-\) , value \ (K +. 1 \) , and the maximum number.

\ (| n | \ 10 ^ {100000} \) .

solution

Category talk.

If \ (n-\) is negative, the first to find a low-order bit is not from \ (9 \) position, to the position \ (1 + \) can. If all the bits are \ (9 \ ) , at the top make up \ (1 \) .

If \ (n-\) non-negative, to find a position to consider:

• This position may be \ (- 1 \) (i.e., the number of this position is not \ (0 \) ).
• The numbers to the right position and that position can be \ (+ 2 \) .

As long as you can find such a position, you can find a positive number to meet the meaning of the questions ...

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