BZOJ 1058 ZJOI 2007 报表统计 平衡树

题目大意：给出一个序列，有几个操作，询问相邻两个数的差值的绝对值的最小值，排序后差值绝对值的最小值。

思路：简单用平衡树或者set水一下就行了。

（我个沙茶最开始sort_min还用的set维护

CODE：

#include <set>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 500010
#define INF 0xf3f3f3f
using namespace std;

int cnt,asks;
int src[MAX],last[MAX];

multiset<int> sorted,list_min;
int sort_min = INF;
char s[20];

int main()
{
    cin >> cnt >> asks;
    for(int i = 1; i <= cnt; ++i)
        scanf("%d",&src[i]),last[i] = src[i],sorted.insert(src[i]);
    for(int i = 2; i <= cnt; ++i)
        list_min.insert(abs(src[i] - src[i - 1]));
    multiset<int>::iterator l;
    for(multiset<int>::iterator it = sorted.begin(); it != sorted.end(); ++it) {
        if(it != sorted.begin())
            sort_min = min(sort_min,abs(*it - *l));
        l = it;
    }
    for(int x,y,i = 1; i <= asks; ++i) {
        scanf("%s",s);
        if(s[0] == 'I') {
            scanf("%d%d",&x,&y);
            multiset<int>::iterator it = sorted.insert(y);
            it--,sort_min = min(sort_min,abs(y - *it)),it++;
            it++,sort_min = min(sort_min,abs(y - *it)),it--;
            it = list_min.find(abs(last[x] - src[x + 1]));
            list_min.erase(it);
            list_min.insert(abs(y - last[x]));
            list_min.insert(abs(src[x + 1] - y));
            last[x] = y;
        }
        else if(s[4] == 'S')    printf("%d\n",sort_min);
        else    printf("%d\n",*list_min.begin());
    }
    return 0;
}