noip模拟赛 #3

T1

给一个环，每个点有一个权值，把环分成三段，求最小的那段的最大值

sol：暴力

二分答案，chk就是把环搞成三倍链，每次枚举起点，后面三个切割点都可以二分找

然后就Rua过去了

//yyc wenle
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = ;
inline LL read()
{
    int x = ,f = ;char ch = getchar();
    for(;!isdigit(ch);ch = getchar())if(ch == '-')f = -f;
    for(;isdigit(ch);ch = getchar())x =  * x + ch - '';
    return x * f;
}
int n;
LL s[ * maxn],a[ * maxn];
LL l,r;
inline LL findnx(int pos,LL x){return upper_bound(s,s +  * n + ,s[pos] + x - ) - s;}
inline int chk(LL x)
{
    int pos;
    for(int i=;i<n;i++)
    {
        pos = i;
        pos = findnx(pos,x);if(pos > i + n)continue;
        pos = findnx(pos,x);if(pos > i + n)continue;
        pos = findnx(pos,x);if(pos > i + n)continue;
        return ;
    }
    return ;
}
int main()
{
    n = read();
    for(int i=;i<=n;i++){a[i] = read();a[n + i] = a[i];a[ * n + i] = a[i];}
    for(int i=;i<= * n;i++)s[i] = s[i - ] + a[i];
    l = ,r = s[n] / ;LL ans;
    while(l <= r)
    {
        LL mid = (l + r) >> ;
        if(chk(mid))l = mid + ,ans = mid;
        else r = mid - ;
    }
    printf("%lld\n",ans);
}
/*
30
1
34
44
13
30
1
9
3
7
7
20
12
2
44
6
9
44
31
17
20
33
18
48
23
19
31
24
50
43
15
*/

T2

树上选出k个点，如果选一个点，也要选他的祖先，默认选0，每个人有一个战斗力和一个花费

求选出的最大战斗力除以最大花费

sol：

分数规划，每个点权变成了zdl - mid * hf

然后就是“树上选出若干点点权大于0”

然后，我们就想歪了

考虑选出的肯定是很多条链，树上选出很多链？那岂不是...

九省联考_林可卡特树

然后想了半天带权二分...咳咳

然后就去想T3了

写完T3才发现这tm不是个树背包吗

然后算算复杂度

小数点后3位，最大10000，那就是1e8

log1e8 * O(n^2)显然挂了

所以需要一个常数小的写法

考虑dfs序，我们选一个点，可以转移到他dfs序后一个点

不选一个点，就转出这棵子树

然后常数非常的优秀，不需要“证明复杂度”和size写法

（学弟预处理size然后T了

//yyc wenle
#include<bits/stdc++.h>
#define LL long long
#define DB long double
using namespace std;
const int maxn = ;
const double inf = 1e9;
inline int read()
{
    int x = ,f = ;char ch = getchar();
    for(;!isdigit(ch);ch = getchar())if(ch == '-')f = -f;
    for(;isdigit(ch);ch = getchar())x =  * x + ch - '';
    return x * f;
}
int n,k;
double s[maxn],p[maxn];
int first[maxn],to[maxn],nx[maxn],cnt;
int dfin[maxn],pos[maxn],dfout[maxn],dfn;
double f[][];int ccnt = ;
inline void add(int u,int v)
{
    to[++cnt] = v;
    nx[cnt] = first[u];
    first[u] = cnt;
}
inline void dfs(int x)
{
    dfin[x] = ++dfn;
    pos[dfn] = x;
    for(int i=first[x];i;i=nx[i])dfs(to[i]);
    dfout[dfin[x]] = dfn;
}
inline int chk(double x)
{
    for(int i=;i<=n + ;i++)
        for(int j=;j<=k + ;j++)
            f[i][j] = -inf;
    f[][] = ;
    for(int i=;i<=n+;i++)
        for(int j=;j<=k+;j++)
        {
            if(f[i][j] == -inf)continue;
            double cur = p[pos[i]] - x * (s[pos[i]]);
            f[dfout[i] + ][j] = max(f[dfout[i] + ][j],f[i][j]);
            f[dfout[i] + ][j + ] = max(f[dfout[i] + ][j + ],f[i][j] + cur);
            for(int xx=first[pos[i]];xx;xx = nx[xx])
            {
                int targ = to[xx];
                f[dfin[targ]][j + ] = max(f[dfin[targ]][j + ],f[i][j] + cur);
                //ccnt++;
            }
        }
    return f[n + ][k + ] >= 0.0;
}
const double eps = 1e-;
int main()
{
    //freopen("rantree.in","r",stdin);
    //freopen("1.txt","r",stdin);
    //freopen("gen.out","w",stdout);
    //freopen("mactree.in","r",stdin);
    //freopen("chain.in","r",stdin);
    //freopen("juhua.in","r",stdin);
    k = read(),n = read();
    if(!k){puts("0.000");return ;}
    for(int i=;i<=n;i++)
    {
        scanf("%lf%lf",&s[i],&p[i]);int py = read();
        add(py,i);
    }
    dfs();
    //cout<<pos[1];
    double l = ,r = 10000.0;
    for(int tms = ;tms <= ;tms++)
    {
        if(r - l <= eps)break;
        double mid = (l + r) / 2.0;
        if(chk(mid))l = mid + eps;
        else r = mid - eps;
    }
    //cout<<ccnt<<endl;
    printf("%.3lf",(l + r) / 2.0);
}

T3

给n个门，每个门是一个位运算和一个数，经过这个门就对这个数操作

求1~m经过这些门之后最大的数

sol：

贪心

1.如果这位选0，改了之后变成1，血赚

2.如果这位选1，改了之后变成0，血亏

3.剩下的，选0肯定比选1更小于m

//yyc wenle
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = ;
inline int read()
{
    int x = ,f = ;char ch = getchar();
    for(;!isdigit(ch);ch = getchar())if(ch == '-')f = -f;
    for(;isdigit(ch);ch = getchar())x =  * x + ch - '';
    return x * f;
}
int n,m;
char opt[];
int num;
int main()
{
    n = read(),m = read();
    int x = ( << ) - ,y = ;
    for(int i=;i<=n;i++)
    {
        scanf("%s",opt);num = read();
        if(opt[] == 'A')x &= num,y &= num;
        if(opt[] == 'X')x ^= num,y ^= num;
        if(opt[] == 'O')x |= num,y |= num;
    }
    int a = ,b = ;
    for(int i = ( << );i;i >>= )
    {
        if((a | i) <= m && (x & i) > (y & i))b |= (x & i),a |= i;
        else b |= (y & i);
    }
    printf("%d\n",b);
}

AK了