求救!!!神犇帮我瞅瞅呗。。。未完。。。调了2个半小时线段树,没调出来,大家帮帮我啊!!!

小詹用st表写。

我的思路就是把中间空着的年份设为无限,然后一点点特判就行了。。。然而没出来。。。

[SCOI2007]降雨量

题干:

Description

  我们常常会说这样的话:“X年是自Y年以来降雨量最多的”。它的含义是X年的降雨量不超过Y年,且对于任意
Y<Z<X,Z年的降雨量严格小于X年。例如2002,2003,2004和2005年的降雨量分别为4920,5901,2832和3890,
则可以说“2005年是自2003年以来最多的”,但不能说“2005年是自2002年以来最多的”由于有些年份的降雨量未
知,有的说法是可能正确也可以不正确的。

Input

  输入仅一行包含一个正整数n,为已知的数据。以下n行每行两个整数yi和ri,为年份和降雨量,按照年份从小
到大排列,即yi<yi+1。下一行包含一个正整数m,为询问的次数。以下m行每行包含两个数Y和X,即询问“X年是
自Y年以来降雨量最多的。”这句话是必真、必假还是“有可能”。

Output

  对于每一个询问,输出true,false或者maybe。

Sample Input

6
2002 4920
2003 5901
2004 2832
2005 3890
2007 5609
2008 3024
5
2002 2005
2003 2005
2002 2007
2003 2007
2005 2008

Sample Output

false
true
false
maybe
false

HINT

100%的数据满足:1<=n<=50000, 1<=m<=10000, -10^9<=yi<=10^9, 1<=ri<=10^9

我的凉凉线段树代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(int i = a;i <= n;i++)
#define lv(i,a,n) for(int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}
int z = 0,n,len = 0;
int rain[50005],pl[50005],ans = 0;
int tree[200005],name[200005];
void build(int o,int l,int r)
{
    if(l == r)
    {
        tree[o] = rain[l];
        name[o] = l;
        return;
    }
    int mid = (l + r) >> 1;
    build(o << 1,l,mid);
    build(o << 1 | 1,mid + 1,r);
    if(tree[o << 1] > tree[o << 1 | 1])
    {
        tree[o] = tree[o << 1];
        name[o] = name[o << 1];
    }
    else if(tree[o << 1] < tree[o << 1 | 1])
    {
        tree[o] = tree[o << 1 | 1];
        name[o] = name[o << 1 | 1];
    }
    else
    {
        tree[o] = INF;
//        name[o] = 0;
    }
}
int query(int o,int l,int r,int x,int y)
{
    cout<<l<<" "<<r<<" "<<x<<" "<<y<<" "<<o<<endl;
    if(l == x && r == y)
    {
        ans = name[o];
        cout<<tree[o]<<endl;
        if(tree[o] == INF)
        {
            if(l != r)
            {
                int mid = (l + r) / 2;
                return max(query(o * 2,l,mid,l,mid),query(o * 2,mid + 1,r,mid + 1,r));
            }
            else
            return -333;
        }
        return tree[o];
    }
    int mid = (l + r) >> 1;
    if(y <= mid)
    {
        return query(o * 2,l,mid,x,y);
    }
    else if(mid < x)
        return query(o * 2 + 1,mid + 1,r,x,y);
    else
    {
        return max(query(o * 2,l,mid,x,mid),query(o * 2 + 1,mid + 1,r,mid + 1,y));
    }
}
int judge(int l,int r)
{
    z = 0;
    int x = query(1,1,len,l + 1,r);
    cout<<x<<endl;
    if(x == rain[r] && z == 1)
        return 2;
    else if(x == rain[r])
        return 1;
    else
        return 3;
}
int main()
{
    read(n);
    int x,y;
    duke(i,1,n)
    {
        read(x);read(y);
        if(pl[len] != x - 1 && i != 1)
        {
            int k = pl[len];
            pl[++len] = k + 1;
            rain[len] = INF;
//            cout<<len<<" "<<rain[len]<<endl;
        }
        pl[++len] = x;
        rain[len] = y;
    }
    build(1,1,len);
//    printf("%d %d\n",tree[1],name[1]);
    int m;
    read(m);
    duke(i,1,m)
    {
        read(x);read(y);
        int p = lower_bound(pl + 1,pl + len + 1,x) - pl;
        int q = lower_bound(pl + 1,pl + len + 1,y) - pl;
        cout<<p<<" "<<q<<endl;
        int t = judge(p,q);
        if(t == 1)
        {
            printf("true\n");
        }
        else if(t == 2)
        {
            printf("maybe\n");
        }
        else
        {
            printf("false\n");
        }
    }
    return 0;
}
/*
6
2002 4920
2003 5901
2004 2832
2005 3890
2007 5609
2008 3024
5
2002 2005
2003 2005
2002 2007
2003 2007
2005 2008
*/
/*
6
2002 4920
2003 5901
2004 2832
2005 3890
2007 5609
2008 3024
2
2002 2007
2003 2007
*/

小詹的st表算法:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<stack>
#include<queue>
#include<vector>
#include<cctype>
using namespace std;
#define space putchar(' ')
#define enter puts("")
#define Mem(a) memset(a, 0, sizeof(a))
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const  db eps = 1e-8;
const int maxn = 5e4 + 5;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = (ans << 3) + (ans << 1) + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

int n, m, a[maxn], yr[maxn];

int dp[maxn][25], b[maxn];
void rmq()
{
    for(int i = 1; i <= n; ++i) dp[i][0] = a[i];
    for(int j = 1; (1 << j) <= n; ++j)
        for(int i = 1; i + (1 << j) - 1 <= n; ++i)
            dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
    int x = 0;
    for(int i = 1; i <= n; ++i)
    {
        b[i] = x;
        if((1 << (x + 1)) <= i + 1) x++;
    }
}
int query(int L, int R)
{
    int k = b[R - L + 1];
    return max(dp[L][k], dp[R - (1 << k) + 1][k]);
}

int main()
{
    n = read();
    for(int i = 1; i <= n; ++i) yr[i] = read(), a[i] = read();
    rmq();
    m = read();
    for(int i = 1; i <= m; ++i)
    {
        int y = read(), x = read();
        if(x <= y) printf("false\n");
        else
        {
            int L = lower_bound(yr + 1, yr + n + 1, y) - yr;
            int R = lower_bound(yr + 1, yr + n + 1, x) - yr;
            bool fl = yr[L] == y, fr = yr[R] == x;
            int ans = 0;
            if(L + (fl ? 1 : 0) <= R - 1) ans = query(L + (fl ? 1 : 0), R - 1);    //别忘了这个判断啊…… 
            if((fr && ans >= a[R]) || (fl && ans >= a[L]) || (fl && fr && (a[L] < a[R] || ans >= a[R]))) printf("false\n");
            else if(R - L != yr[R] - yr[L] || !fl || !fr) printf("maybe\n");
            else printf("true\n");
        }
    }
    return 0;
}

 

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本文链接:https://www.cnblogs.com/DukeLv/p/9526704.html