51nod1287(二分/线段树区间最值&单点更新)

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1287

题意：中文题诶～

解法１：b[i] 存储 max(a[0], ....., a[i])，显然 b 是单调不减的，所以直接二分 x，再更新 a 和 b 数组即可；

代码：

#include <iostream>
#include <stdio.h>
using namespace std;

const int MAXN = 5e4 + 10;
int a[MAXN], b[MAXN], x;

int main(void){
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i++){
        scanf("%d", &a[i]);
        if(i == 0) b[i] = a[i];
        else b[i] = max(b[i - 1], a[i]);
    }
    while(m--){
        scanf("%d", &x);
        if(x <= a[0] || x > b[n - 1]) continue;
        int pos = lower_bound(b, b + n, x) - b;
        a[pos - 1]++;
        b[pos - 1] = max(b[pos - 1], a[pos - 1]);
    }
    for(int i = 0; i < n; i++){
        printf("%d
", a[i]);
    }
    return 0;
}

解法２：用线段树维护区间最大值，再更新一下 a 数组即可；

代码：

#include <iostream>
#include <stdio.h>
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
using namespace std;

const int MAXN = 5e4 + 10;
int Max[MAXN << 2];
int a[MAXN];

void push_up(int rt){
    Max[rt] = max(Max[rt << 1], Max[rt << 1 | 1]);
}

void build(int l, int r, int rt){
    if(l == r){
        Max[rt] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    push_up(rt);
}

void update(int p, int x, int l, int r, int rt){
    if(l == r){
        Max[rt] += x;
        return;
    }
    int mid = (l + r) >> 1;
    p <= mid ? update(p, x, lson) : update(p, x, rson);
    push_up(rt);
}


int query(int x, int l, int r, int rt){
    if(l == r) return l;
    int mid = (l + r) >> 1;
    return Max[rt << 1] >= x ? query(x, lson) : query(x, rson);
}


int main(void){
    int n, m, x;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
    }
    build(1, n , 1);
    while(m--){
        scanf("%d", &x);
        if(x > Max[1] || x <= a[1]) continue;
        int index = query(x, 1, n, 1);
        a[index - 1] += 1;
        update(index - 1, 1, 1, n, 1);
    }
    for(int i = 1; i <= n; i++){
        printf("%d
", a[i]);
    }
    return 0;
}