7.24模拟练补题

HDU 5573 Binary Tree 

获得姿势：要构造数，想到用二进制构造。

一直向最左路构造，最左路的和是（2^k）-1,加入给定n,那么被减去的数字是d = （2^k)-1-n.  当我要减去一个数字时，从上往下走，对于总和相当于减去二倍的数字，所以要d/2.但前提是保证d为偶数。即为奇数时d = (2^k)-n.因为减去一个数，总是损失二倍的。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
int main()
{
    int t,kase = 0;
    scanf("%d",&t);
    while(t--)
    {
        ll n,k;
        scanf("%I64d %I64d",&n,&k);
        printf("Case #%d:\n",++kase);
        ll sum = (1LL<<k)-1; // - 和 + 优先级比<<高
        int flag = 0;
        sum -= n;
        if(sum%2)
        {
            sum += 1;
            flag = 1;
        }
        sum /= 2;
        for(int i = 0; i<k-1; i++)
        {
            printf("%I64d ",1LL<<i);
            if(sum&(1LL<<i)) printf("-\n");
            else printf("+\n");
        }
        if(flag) printf("%I64d +\n",(1LL<<(k-1))+1);
        else printf("%I64d +\n",1LL<<(k-1));
    }
    return 0;
}

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 HDU 5521 Meeting

先把自己的SB代码保存下来，昨天WA了十五发，最近应该不会再看了。QAQ

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
typedef long long ll;
const int maxn = 3e6+5;
ll d1[maxn],d2[maxn];
bool done[maxn];
const ll INF = 0x3f3f3f3f3f3f3f3fLL;
struct edge
{
    int to;
    ll cost;
    friend bool operator < (edge A,edge B)
    {
        return A.cost>B.cost;
    }
};
vector<edge> g[maxn];
vector<int> dd;
void add(int u,int v,ll w)
{
    g[u].push_back((edge){v,w});
 //   g[v].push_back((edge){u,w});
}
void dijkstra(int s,int e,ll d[])
{
    priority_queue<edge> q;
    memset(done,false,sizeof(done));
    for(int i = 1; i<=maxn; i++) d[i] = INF;
    d[s] = 0;
    q.push((edge){s,0});
    while(!q.empty())
    {
        edge cur = q.top();
        q.pop();
        int v = cur.to;
        if(done[v]) continue;
        done[v] = true;
        for(int i = 0; i<g[v].size(); i++)
        {
            cur = g[v][i];
            if(d[cur.to]>d[v]+cur.cost)
            {
                d[cur.to] =  d[v]+cur.cost;
                q.push((edge){cur.to,d[cur.to]});
            }
        }
    }
}
int main()
{
    int t,kase = 0;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        ll tt;
        int s,x;
        int in,out;
        cin>>n>>m;
        for(int i = 1; i<=m; i++)
        {
            scanf("%I64d %d",&tt,&s);
            in = n+i;
            out = n+m+i;
            add(in,out,tt);
          //  add(out,in,tt);
            for(int j = 1; j<=s; j++)
            {
                scanf("%d",&x);
                add(x,in,0);
                add(out,x,0);
            }
        }
        dijkstra(1,n,d1);
        dijkstra(n,1,d2);
        ll minn = INF;
        printf("Case #%d: ",++kase);
        for(int i = 1; i<=n; i++) minn = min(minn,max(d1[i],d2[i]));
        if(minn == INF)
        {
             printf("Evil John\n");
             continue;
        }
        printf("%I64d\n",minn);
        for(int i = 1; i<=n; i++)
            if(max(d1[i],d2[i])==minn)
        {
            dd.push_back(i);
        }
        int first = 0;
        for(int i=0;i<dd.size();i++)
        {
            if(first==0)
            {
                printf("%d",dd[i]);
                first = 1;
            }
            else
                printf(" %d",dd[i]);
        }
        printf("\n");
        for(int i = 1; i<=maxn; i++) g[i].clear();
        dd.clear();
    }
    return 0;
}

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HDU 5441 Travel

呀，最近自己有毒，不wa上十几发是不会过的，不对，wa了十几发也不一定过。

对边排序，对查询排序，并查集。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1e5+5;
const int maxn1 = 2e4+5;
struct edge
{
    int u,v,w;
};
edge e[maxn];
int pre[maxn1];
int cnt[maxn1];
int n,m,qq;
struct query
{
    int k;
    int ans;
    int x;
};
query q[5005];

void inin()
{
    for(int i=1;i<=n;i++)
    {
        pre[i] = i;
        cnt[i] = 1;
    }
}
int find1(int x)
{
    int r = x;
    while(r!=pre[r])
    {
        r = pre[r];
    }
    int i = x,j;
    while(pre[i]!=r)
    {
        j = pre[i];
        pre[i] = r;
        i = j;
    }
    return r;
}
bool cmp1(edge A,edge B)
{
    return A.w<B.w;
}

bool cmp2(query A,query B)
{
    return A.x<B.x;
}

bool cmp3(query A,query B)
{
    return A.k<B.k;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(e,0,sizeof(e));
        memset(q,0,sizeof(q));
        cin>>n>>m>>qq;
        int u,v,w;
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].w);
        }
        sort(e+1,e+m+1,cmp1);
        int x;
        for(int i=1;i<=qq;i++)
        {
            scanf("%d",&x);
            q[i].k = i;
            q[i].x = x;
        }
        sort(q+1,q+qq+1,cmp2);
        int index = 1;
        inin();
        int fx = 0;
        int fy = 0;
        int ans = 0;
        for(int i=1;i<=qq;i++)   //手残，加了个index<=m,导致后面的不更新了
        {
            while(e[index].w<=q[i].x&&index<=m)
            {
                fx = find1(e[index].u);
                fy = find1(e[index].v);
                if(fx != fy)
                {
                    pre[fx] = fy;
                    ans -= (cnt[fx]*(cnt[fx]-1));
                    ans -= (cnt[fy]*(cnt[fy]-1));
                    cnt[fy] += cnt[fx];
                    ans += (cnt[fy]*(cnt[fy]-1));
                }
                index++;

            }q[i].ans = ans;

        }
        sort(q+1,q+qq+1,cmp3);
        for(int i=1;i<=qq;i++) printf("%d\n",q[i].ans);
    }
    return 0;
}

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转载于:https://www.cnblogs.com/littlepear/p/5700194.html