G - Multidimensional Queries 线段树结合状压

别说谁变了你拦得住时间么 提交于 2019-12-01 05:00:23

题意 给出一个k维度空间,求[l,r]的最大曼哈顿距离。
用二进制枚举绝对值相加可能出现的所有情况。用线段树来维护单点修改,区间查询。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
    char ch = getchar(); ll x = 0, f = 1;
    while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
struct node { ll s, t, d, w; bool operator<(const node &a)const { return w == a.w ? d < a.d : w < a.w; }}a[N];
bool cmp(node &a, node &b)
{
    return a.s < b.s;
}
void minx(ll &x, ll y) { if (y == -1)return; x = x == -1 ? y : min(x, y); }
int n, m,k;
ll dp[N][205];
int main()
{
    n = read(), m = read(), k = read();
    ll s, t, d, w;
    up(i, 0, k)
    {
        s = read(), t = read(), d = read(), w = read();
        a[i] = node{ s,t,d,w };
    }
    sort(a, a + k, cmp);
    memset(dp, -1, sizeof(dp));
    dp[1][0] = 0;
    priority_queue<node>q;
    int cnt = 0;
    upd(i, 1, n)
    {
        while (cnt < k&&a[cnt].s <= i)q.push(a[cnt]),cnt++;
        while (!q.empty() && q.top().t < i)q.pop();
        if (q.empty())
        {
            upd(j, 0, m) minx(dp[i + 1][j], dp[i][j]);
            continue;
        }
        node temp = q.top();
        upd(j, 0, m)
        {
            if (dp[i][j] == -1)continue;
            minx(dp[temp.d + 1][j], dp[i][j] + temp.w);
            if (j < m)minx(dp[i + 1][j + 1], dp[i][j]);
        }
    }
    ll ans = 1e18;
    upd(i, 0, m)ans = min(ans, dp[n + 1][i]);
    printf("%lld", max(0ll, ans));
}
标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!