Codeforces Round #600 (Div. 2)

C - Sweets Eating

基本思路：对于每个i，选出最小的前i个值，越大的值放在越前面，即可得到最小的penalty

假设dp[i]表示前i个元素能获得的最小penalty，那么dp[i + m] = dp[i] + sum[1 : i+m]，即[i+1, i+m]放在第一天，其他的以此往后移一天

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+7;
const ll mod = 998244353;
#define afdafafafdafaf y1;
int ar[maxn], n, m;
 
ll p[maxn], s[maxn];
int main()
{
    scanf("%d%d", &n, &m);
    for(int i=1;i<=n;i++)scanf("%d", ar+i);
    sort(ar+1, ar+n+1);
    for(int i=1;i<=n;i++)p[i] = p[i-1] + ar[i];
    //for(int i=1;i<=n;i++)printf("%lld%c", p[i], i==n?'
' : ' ');
    for(int i=1;i<=m;i++){
        ll ins = i, x = p[i];
        while(ins <= n){
            s[ins] = x;
            ins += m;
            if(ins <= n)x += p[ins];
        }
    }
    for(int i=1;i<=n;i++)printf("%lld%c", s[i], i==n?'
' : ' ');
    return 0;
}

D - Harmonious Graph

边排序+并查集

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+7;
const ll mod = 998244353;
#define afdafafafdafaf y1;
int ar[maxn], n, m;
 
pair<int,int> p[maxn];
int f[maxn];
int findx(int x){
    return f[x] == x ? x : f[x] = findx(f[x]);
}
int main()
{
    scanf("%d%d", &n, &m);
    for(int i=1;i<=n;i++)f[i] = i;
    for(int i=1;i<=m;i++){
        int a,b;scanf("%d%d", &a, &b);
        if(a > b)swap(a, b);
        p[i].first = a;
        p[i].second = b;
        a = findx(a);
        b = findx(b);
        f[a] = b;
    }
    sort(p + 1, p + m + 1);
    int pa = 0, pb = 0;
    p[m+1].first = int(1e9);
    //cout<<p[m+1].first<<'
';
    int ans=0;
    for(int i=1;i<=m+1;i++){
        //cout<<"i = " << i << ' ' << p[i].first<<" "<<p[i].second<<"
";
        if(p[i].first > pb){
            int x = pa;
            //cout<<pa<<" "<<pb<<"
";
            for(int j=pa+1;j<=pb;j++){
                x = findx(x);
                int y = j;
                y = findx(y);
                if(x != y){
                    f[x] = y;
                    ans++;
                }
            }
            pa = p[i].first;
            pb = p[i].second;
        }
        else{
            pb = max(pb, p[i].second);
        }
    }
    printf("%d
", ans);
    return 0;
}

E - Antenna Coverage

按照每个区间的头尾排序

正序遍历，求出覆盖前i个节点所需要的最小花费，答案即que(m)

因为两段如果要相接的话，前一段的头一定在后一段的头的前面，所以可以证明排序后正序遍历是正确的。

（比赛的时候求了前后缀，遍历相加，所以代码里是que_l，忽略即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+7;
const ll mod = 998244353;
#define afdafafafdafaf y1;
int ar[maxn], n, m;
 
int x[maxn], s[maxn], l[maxn], r[maxn];
ll lp[maxn], rp[maxn];
pair<int,int> p[maxn];
int lmn[maxn << 2], rmn[maxn << 2];
void set_l(int x, int val){
    while(x > 0){
        if(val >= lmn[x])break;
        lmn[x] = min(lmn[x], val);
        x -= x & (-x);
        //cout<<"x_set = " << x << '
';
    }
}
int que_l(int x){
    if(x == 0)return 0;
    int ans = m;
    while(x <= m){
        ans = min(ans, lmn[x]);
        x += x & (-x);
        //cout<<"x_que = " << x << '
';
    }
    return ans;
}
 
int main()
{
    scanf("%d%d", &n, &m);
    for(int i=1;i<=n;i++)scanf("%d%d", x + i, s + i);
    for(int i=1;i<=n;i++){
        int a = x[i], b = s[i];
        p[i].first = max(1, a - b);
        p[i].second = min(m, a + b);
    }
    sort(p+1, p+n+1);
    for(int i=1;i<=n;i++){
        l[i] = p[i].first;
        r[i] = p[i].second;
    }
    for(int i = 0; i < (maxn << 2); i++)lmn[i] = rmn[i] = m;
    set_l(0, 0);
    for(int i=1;i<=n;i++){
        int a = l[i], b = r[i], ins = 0;
        while(a >= 1 || b <= m){
            int l_a = que_l(a - 1);
            set_l(b, l_a + ins);
            if(a == 1 && b == m)break;
            if(a > 1)a--;
            if(b < m)b++;
            ins++;
        }
    }
    printf("%d
", que_l(m));
    return 0;
}

F - Cheap Robot

不会