11.6 校内模拟赛解题报告

T1

考虑没个点应该向左或者向右传多少，或者它自身接受左边传来了多少，右边传来了多少，因为一次只能传一个，所以传送的多少就是时间。
对所有的这些情况取 max。

/*
Date:
Source:
Knowledge: 
*/
#include <iostream>
#include <cstdio>
#include <cmath>
#define orz cout << "AK IOI" << "\n"
#define int long long

using namespace std;
const int maxn = 1e5 + 10;

int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch <= '9' && ch >= '0') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void print(int X)
{
	if(X < 0) X = ~(X - 1), putchar('-');
	if(X > 9) print(X / 10);
	putchar(X % 10 ^ '0');
}
int Max(int a, int b){
	return a > b ? a : b;
}
int Min(int a, int b){
	return a < b ? a : b;
}
int n, ans, a[maxn], sum[maxn];
signed main()
{
	//freopen("link.in", "r", stdin);
	//freopen("link.out", "w", stdout);
	n = read();
	for(int i = 1; i <= n; i++) a[i] = read(), sum[i] = sum[i - 1] + a[i];
	int tmp = sum[n] / n;
    for(int i = 1; i <= n; i++)
    {
        int tmp1 = sum[i - 1] - (tmp * (i - 1)); //左边的
		int tmp2 = sum[n] - sum[i] - tmp * (n - i); // 右边的
		if(tmp1 < 0 && tmp2 < 0) ans = Max(ans, abs(tmp1 + tmp2));
		else ans = Max(ans, Max(abs(tmp1), abs(tmp2))); 
    }
    print(ans);
	//fclose(stdin);
	//fclose(stdout);
	return 0;
}

T2

答案与城市的顺序无关，题目要求取走的数必须是所有已经取走的数的倍数或者约数。对数组进行排序，转化成取倍数的问题，用类似于埃氏筛的方法求解。

/*
Date:
Source:
Knowledge:
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#define orz cout << "AK IOI" << "\n"

using namespace std;
const int maxn = 1e6 + 10;

int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch <= '9' && ch >= '0') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void print(int X)
{
	if(X < 0) X = ~(X - 1), putchar('-');
	if(X > 9) print(X / 10);
	putchar(X % 10 ^ '0');
}
int Max(int a, int b){
	return a > b ? a : b;
}
int Min(int a, int b){
	return a < b ? a : b;
}
int n, ans, a[maxn], f[maxn], cnt[maxn]; 

int main()
{
	//freopen("pigeon.in", "r", stdin);
	//freopen("pigeon.out", "w", stdout);
	n = read();
	for(int i = 1; i <= n; i++) a[i] = read(), f[a[i]]++, cnt[a[i]]++;
	sort(a + 1, a + n + 1);
	for(int i = 1; i <= maxn; i++)
	{
		if(f[i])
		{
			ans = Max(ans, f[i]);
			for(int j = i + i; j <= maxn; j += i)
				if(cnt[j]) f[j] = Max(f[j], f[i] + cnt[j]);
		}
	}
	print(ans);
	fclose(stdin);
	fclose(stdout);
	return 0;
}

T3

二分答案。
每次检验的时候会发现，一组卡牌中的一张选了会导致其他某组卡牌必须选另外一张，从而转化为 2-SAT 问题。
对于 所有卡牌战斗力的最大值减去最小值小于等于 100 的部分分，输出 \(0\)。
由于数据范围过大，无法直接 2-SAT 建图，注意到所有的边都是在一个区间里面的边，所以用线段树优化建图。

贴一下 std 吧。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<algorithm>

using namespace std;

const int BUF_SIZE = 30;
char buf[BUF_SIZE], *buf_s = buf, *buf_t = buf + 1;
  
#define PTR_NEXT() \
    { \
        buf_s ++; \
        if (buf_s == buf_t) \
        { \
            buf_s = buf; \
            buf_t = buf + fread(buf, 1, BUF_SIZE, stdin); \
        } \
    }
   
#define readint(_n_) \
    { \
        while (*buf_s != '-' && !isdigit(*buf_s)) \
            PTR_NEXT(); \
        bool register _nega_ = false; \
        if (*buf_s == '-') \
        { \
            _nega_ = true; \
            PTR_NEXT(); \
        } \
        int register _x_ = 0; \
        while (isdigit(*buf_s)) \
        { \
            _x_ = _x_ * 10 + *buf_s - '0'; \
            PTR_NEXT(); \
        } \
        if (_nega_) \
            _x_ = -_x_; \
        (_n_) = (_x_); \
    }

#define readstr(_s_) \
    { \
        while (!isupper(*buf_s)) \
            PTR_NEXT(); \
        char register *_ptr_ = (_s_); \
        while (isupper(*buf_s) || *buf_s == '-') \
        { \
            *(_ptr_ ++) = *buf_s; \
            PTR_NEXT(); \
        } \
        (*_ptr_) = '\0'; \
    }

#define readlonglong(_n_) \
    { \
        while (*buf_s != '-' && !isdigit(*buf_s)) \
            PTR_NEXT(); \
        bool register _nega_ = false; \
        if (*buf_s == '-') \
        { \
            _nega_ = true; \
            PTR_NEXT(); \
        } \
        long long register _x_ = 0; \
        while (isdigit(*buf_s)) \
        { \
            _x_ = _x_ * 10 + *buf_s - '0'; \
            PTR_NEXT(); \
        } \
        if (_nega_) \
            _x_ = -_x_; \
        (_n_) = (_x_); \
    }

#define wmt 1,(n<<1),1
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int maxn=200010;
const int maxp=maxn+(maxn<<2);
const int maxm=maxn+maxp+maxn*40;

int n,size,cnt,en,t,dfn[maxp],low[maxp],s[maxp],belong[maxp],pos[maxn];

bool instack[maxp];

struct edge
{
	int e;
	edge *next;
}*v[maxp],ed[maxm];

void add_edge(int s,int e)
{
	en++;
	ed[en].next=v[s];v[s]=ed+en;v[s]->e=e;
}

struct rec
{
	int v,p;
	rec(){}
	rec(int a,int b)
	{
		v=a;p=b;
	}
}z[maxn];

bool operator<(const rec &a,const rec &b)
{
	return a.v<b.v;
}

void dfs(int p)
{
	t++;
	dfn[p]=low[p]=t;
	instack[p]=true;
	s[++size]=p;
	for (edge *e=v[p];e;e=e->next)
		if (!dfn[e->e])
		{
			dfs(e->e);
			low[p]=min(low[p],low[e->e]);
		}
		else
		{
			if (instack[e->e]) low[p]=min(low[p],dfn[e->e]);
		}
	if (dfn[p]==low[p])
	{
		cnt++;
		while (s[size]!=p)
		{
			belong[s[size]]=cnt;
			instack[s[size]]=false;
			size--;
		}
		belong[p]=cnt;
		instack[p]=false;
		size--;
	}
}

void build(int l,int r,int rt)
{
	if (l==r)
	{
		add_edge(rt+(n<<1),z[l].p<=n?z[l].p+n:z[l].p-n);
		return;
	}
	int m=(l+r)>>1;
	build(lson);
	build(rson);
	add_edge(rt+(n<<1),(rt<<1)+(n<<1));
	add_edge(rt+(n<<1),(rt<<1|1)+(n<<1));
}

void insert(int l,int r,int rt,int nowl,int nowr,int p)
{
	if (nowl<=l && r<=nowr)
	{
		add_edge(p,rt+(n<<1));
		return;
	}
	int m=(l+r)>>1;
	if (nowl<=m) insert(lson,nowl,nowr,p);
	if (m<nowr) insert(rson,nowl,nowr,p);
}

bool check(int k)
{
	en=0;cnt=0;
	memset(v,0,sizeof(v));
	memset(dfn,0,sizeof(dfn));

	build(wmt);

	int r=1,l=1;

	for (int a=1;a<=(n<<1);a++)
	{
		int op,p=z[a].p;
		if (p<=n) op=pos[p+n];
		else op=pos[p-n];
		while (r<=a && z[r].v <= z[a].v-k)
			r++;
		if (r<a && r>=1 && z[r].v > z[a].v-k) 
		{
			if (op>=r && op<=a-1)
			{
				if (op>r) insert(wmt,r,op-1,z[a].p);
				if (op<a-1) insert(wmt,op+1,a-1,z[a].p);
			}
			else insert(wmt,r,a-1,z[a].p);
		}
		while (l<=(n<<1) && z[l].v < z[a].v+k)
			l++;
		l--;
		if (l>a && l<=(n<<1) && z[l].v < z[a].v+k) 
		{
			if (op>=a+1 && op<=l)
			{
				if (op>a+1) insert(wmt,a+1,op-1,z[a].p);
				if (op<l) insert(wmt,op+1,l,z[a].p);
			}
			else insert(wmt,a+1,l,z[a].p);
		}
	}

	for (int a=1;a<=(n<<1);a++)
		if (!dfn[a]) dfs(a);
	for (int a=1;a<=n;a++)
		if (belong[a]==belong[a+n]) return false;
	return true;
}

int main()
{
	readint(n);
	int minv=0x3f3f3f3f,maxv=-0x3f3f3f3f;
	int x=0;
	for (int a=1;a<=n;a++)
	{
		int v1,v2;
		readint(v1);
		readint(v2);
		z[++x]=rec(v1,a);
		z[++x]=rec(v2,a+n);
		minv=min(minv,min(v1,v2));
		maxv=max(maxv,max(v1,v2));
	}
	if (maxv-minv+1 < n)
	{
		printf("0\n");
		return 0;
	}
	sort(z+1,z+x+1);
	for (int a=1;a<=(n<<1);a++)
		pos[z[a].p]=a;
	int l=0,r=1000000001;
	while (l+1!=r)
	{
		int m=(l+r)>>1;
		if (check(m)) l=m;
		else r=m;
	}
	printf("%d\n",l);

	return 0;
}