Codeforces Round #600 (Div. 2)
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 2e5+7; 5 const ll mod = 998244353; 6 #define afdafafafdafaf y1; 7 int ar[maxn], n, m; 8 9 ll p[maxn], s[maxn]; 10 int main() 11 { 12 scanf("%d%d", &n, &m); 13 for(int i=1;i<=n;i++)scanf("%d", ar+i); 14 sort(ar+1, ar+n+1); 15 for(int i=1;i<=n;i++)p[i] = p[i-1] + ar[i]; 16 //for(int i=1;i<=n;i++)printf("%lld%c", p[i], i==n?'\n' : ' '); 17 for(int i=1;i<=m;i++){ 18 ll ins = i, x = p[i]; 19 while(ins <= n){ 20 s[ins] = x; 21 ins += m; 22 if(ins <= n)x += p[ins]; 23 } 24 } 25 for(int i=1;i<=n;i++)printf("%lld%c", s[i], i==n?'\n' : ' '); 26 return 0; 27 }
边排序+并查集
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 2e5+7; 5 const ll mod = 998244353; 6 #define afdafafafdafaf y1; 7 int ar[maxn], n, m; 8 9 pair<int,int> p[maxn]; 10 int f[maxn]; 11 int findx(int x){ 12 return f[x] == x ? x : f[x] = findx(f[x]); 13 } 14 int main() 15 { 16 scanf("%d%d", &n, &m); 17 for(int i=1;i<=n;i++)f[i] = i; 18 for(int i=1;i<=m;i++){ 19 int a,b;scanf("%d%d", &a, &b); 20 if(a > b)swap(a, b); 21 p[i].first = a; 22 p[i].second = b; 23 a = findx(a); 24 b = findx(b); 25 f[a] = b; 26 } 27 sort(p + 1, p + m + 1); 28 int pa = 0, pb = 0; 29 p[m+1].first = int(1e9); 30 //cout<<p[m+1].first<<'\n'; 31 int ans=0; 32 for(int i=1;i<=m+1;i++){ 33 //cout<<"i = " << i << ' ' << p[i].first<<" "<<p[i].second<<"\n"; 34 if(p[i].first > pb){ 35 int x = pa; 36 //cout<<pa<<" "<<pb<<"\n"; 37 for(int j=pa+1;j<=pb;j++){ 38 x = findx(x); 39 int y = j; 40 y = findx(y); 41 if(x != y){ 42 f[x] = y; 43 ans++; 44 } 45 } 46 pa = p[i].first; 47 pb = p[i].second; 48 } 49 else{ 50 pb = max(pb, p[i].second); 51 } 52 } 53 printf("%d\n", ans); 54 return 0; 55 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 2e5+7; 5 const ll mod = 998244353; 6 #define afdafafafdafaf y1; 7 int ar[maxn], n, m; 8 9 int x[maxn], s[maxn], l[maxn], r[maxn]; 10 ll lp[maxn], rp[maxn]; 11 pair<int,int> p[maxn]; 12 int lmn[maxn << 2], rmn[maxn << 2]; 13 void set_l(int x, int val){ 14 while(x > 0){ 15 if(val >= lmn[x])break; 16 lmn[x] = min(lmn[x], val); 17 x -= x & (-x); 18 //cout<<"x_set = " << x << '\n'; 19 } 20 } 21 int que_l(int x){ 22 if(x == 0)return 0; 23 int ans = m; 24 while(x <= m){ 25 ans = min(ans, lmn[x]); 26 x += x & (-x); 27 //cout<<"x_que = " << x << '\n'; 28 } 29 return ans; 30 } 31 32 int main() 33 { 34 scanf("%d%d", &n, &m); 35 for(int i=1;i<=n;i++)scanf("%d%d", x + i, s + i); 36 for(int i=1;i<=n;i++){ 37 int a = x[i], b = s[i]; 38 p[i].first = max(1, a - b); 39 p[i].second = min(m, a + b); 40 } 41 sort(p+1, p+n+1); 42 for(int i=1;i<=n;i++){ 43 l[i] = p[i].first; 44 r[i] = p[i].second; 45 } 46 for(int i = 0; i < (maxn << 2); i++)lmn[i] = rmn[i] = m; 47 set_l(0, 0); 48 for(int i=1;i<=n;i++){ 49 int a = l[i], b = r[i], ins = 0; 50 while(a >= 1 || b <= m){ 51 int l_a = que_l(a - 1); 52 set_l(b, l_a + ins); 53 if(a == 1 && b == m)break; 54 if(a > 1)a--; 55 if(b < m)b++; 56 ins++; 57 } 58 } 59 printf("%d\n", que_l(m)); 60 return 0; 61 }
F - Cheap Robot
不会
