• Codeforces 1167E Range Deleting


    Range Deleting

    考虑枚举L, R明显具有可二分性, 然后预处理一些东西, 二分就好啦。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 1e6 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}
    
    int n, x, a[N];
    int mnPos[N], mxPos[N];
    int preMax[N], sufMin[N];
    bool preCan[N], sufCan[N];
    
    inline bool check(int l, int r) {
        if(!preCan[l - 1]) return false;
        if(!sufCan[r + 1]) return false;
        if(sufMin[r + 1] < preMax[l - 1]) return false;
        return true;
    }
    
    int main() {
        memset(mnPos, inf, sizeof(mnPos));
        scanf("%d%d", &n, &x);
        sufMin[x + 1] = inf;
        preMax[0] = -inf;
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++) {
            chkmax(mxPos[a[i]], i);
            chkmin(mnPos[a[i]], i);
        }
        preMax[1] = mxPos[1];
        for(int i = 2; i <= x; i++) preMax[i] = max(preMax[i - 1], mxPos[i]);
        sufMin[x] = mnPos[x];
        for(int i = x - 1; i >= 1; i--) sufMin[i] = min(sufMin[i + 1], mnPos[i]);
        preCan[0] = true;
        for(int i = 1; i <= x; i++) {
            if(!preCan[i - 1]) break;
            if(mnPos[i] > preMax[i - 1]) preCan[i] = true;
        }
        sufCan[x + 1] = true;
        for(int i = x; i >= 1; i--) {
            if(!sufCan[i + 1]) break;
            if(mxPos[i] < sufMin[i + 1]) sufCan[i] = true;
        }
        LL ans = 0;
        for(int i = 1; i <= x; i++) {
            int low = i, high = x, p = i - 1;
            while(low <= high) {
                int mid = low + high >> 1;
                if(!check(i, mid)) p = mid, low = mid + 1;
                else high = mid - 1;
            }
            ans += x - p;
        }
        printf("%lld
    ", ans);
        return 0;
    }
    
    /*
    */
  • 相关阅读:
    .NET Obfuscator Dotfuscator 入门
    查询集合已修改;可能无法执行枚举操作
    泛型(C# 2。0 编程指南) <一>
    在服务器上部署VS 2008 ReportViewer,完美支持中文
    dataGridView 闪烁 和 listview 闪烁 的解决办法。
    Asp.Net 调试客户端脚本
    疑是Microsoft Enterprise Library June 2005的一个小bug (续)
    MagicAjax 0.30版的更新(翻译)
    疑是Microsoft Enterprise Library June 2005的一个小bug
    在web页面中水晶报表显示速度过慢的原因
  • 原文地址:https://www.cnblogs.com/CJLHY/p/10873961.html
Copyright © 2020-2023  润新知