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 + ;
 int a[MAXN], b[MAXN], x;

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

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 + ;
 int Max[MAXN << ];
 int a[MAXN];

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

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

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