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 0xf3f3f3fusing 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;}