Codeforces Round #592 (Div. 2)
求gcd，遍历下p/w向下w个值即可，复杂度O(w)
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e3+7; 5 const ll mod = 1e9 + 9; 6 #define afdafafafdafaf y1; 7 int ar[maxn], n; 8 9 10 void e_gcd(ll a, ll b, ll &gcd, ll &x, ll &y) 11 { 12 if (b==0){ 13 x=1; 14 y=0; 15 gcd=a; 16 }else{ 17 e_gcd(b, a%b, gcd, y, x);//注意要交换x和y 18 y=y-x*(a/b);//有上面推导出来的公式 19 //x1 = y2（等式两边a的系数相同） 20 //y1 = x2 - (a / b) * y2 （等式两边b的系数相同） 21 } 22 } 23 int main() 24 { 25 ll n,p,w,d; 26 scanf("%lld%lld", &n, &p); 27 scanf("%lld%lld", &w, &d); 28 if(n * w < p){ 29 printf("-1\n"); 30 return 0; 31 } 32 ll gcd, x, y; 33 e_gcd(w, d, gcd, x, y); 34 if(p % gcd != 0){ 35 printf("-1\n"); 36 return 0; 37 } 38 w /= gcd; 39 d /= gcd; 40 p /= gcd; 41 ll ins = p / w; 42 for(int i=0;i<=min(w, ins);i++){ 43 if((p - (ins - i) * w) % d == 0){ 44 x = ins - i; 45 y = (p - (ins - i) * w) / d; 46 break; 47 } 48 } 49 if(x >= 0 && y >= 0 && n - x - y >= 0){ 50 printf("%lld %lld %lld\n", x, y, n - x - y); 51 } 52 else printf("-1\n"); 53 return 0; 54 }
初步分析如果出现度大于等于3的点，没有配色方案，那么树一定是一条链，从链的一头开始遍历，dp即可
no[u].a[i][j]值得是u节点涂第i种颜色，他的子节点涂第j种颜色需要的最小花费，dp即可，过程略微繁琐
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e5+7; 5 const ll mod = 1e9 + 9; 6 #define afdafafafdafaf y1; 7 int ar[maxn], n; 8 9 struct node{ 10 ll a[4][4]; 11 }no[maxn]; 12 int arr[4][maxn]; 13 vector<int> g[maxn]; 14 node dfs(int u, int pre){ 15 node nou; 16 memset(nou.a, 0, sizeof(nou.a)); 17 for(int v : g[u]){ 18 if(v == pre)continue; 19 node mid = dfs(v, u); 20 for(int i=1;i<=3;i++){ 21 for(int j=1;j<=3;j++){ 22 if(i != j)nou.a[i][j] += mid.a[i][j]; 23 } 24 } 25 } 26 memset(no[u].a, 0, sizeof(no[u].a)); 27 no[u].a[1][2] = arr[1][u] + nou.a[2][3]; 28 no[u].a[1][3] = arr[1][u] + nou.a[3][2]; 29 no[u].a[2][1] = arr[2][u] + nou.a[1][3]; 30 no[u].a[2][3] = arr[2][u] + nou.a[3][1]; 31 no[u].a[3][1] = arr[3][u] + nou.a[1][2]; 32 no[u].a[3][2] = arr[3][u] + nou.a[2][1]; 33 return no[u]; 34 } 35 int in[maxn], av[maxn]; 36 void dfs2(int u, int pre, int col, ll val, int pre_col){ 37 av[u] = col; 38 if(in[u] == 1 && pre != 0){ 39 //assert(val == arr[col][u]); 40 } 41 for(int v : g[u]){ 42 if(v==pre)continue; 43 ll mn = 1e18; 44 int ins = 0; 45 for(int j=1;j<=3;j++){ 46 if(col != j){ 47 if(mn > no[u].a[col][j] && j != pre_col){ 48 mn = no[u].a[col][j]; 49 ins = j; 50 } 51 } 52 } 53 //assert(val == mn); 54 dfs2(v, u, ins, val - arr[col][u], col); 55 } 56 } 57 int main() 58 { 59 scanf("%d", &n); 60 for(int i=1;i<=3;i++){ 61 for(int j=1;j<=n;j++)scanf("%d", &arr[i][j]); 62 } 63 for(int i=1;i<n;i++){ 64 int a,b;scanf("%d%d", &a, &b); 65 g[a].push_back(b); 66 g[b].push_back(a); 67 in[a]++; 68 in[b]++; 69 } 70 for(int i=1;i<=n;i++){ 71 if(in[i] >= 3){ 72 printf("-1\n"); 73 return 0; 74 } 75 } 76 int be = 0; 77 int xx = 0; 78 for(int i=1;i<=n;i++){ 79 if(in[i] == 1){ 80 be = i; 81 xx++; 82 } 83 else{ 84 assert(in[i] == 2); 85 } 86 } 87 assert(xx == 2); 88 node mid = dfs(be, 0); 89 ll ans = 1e18, ins = 0; 90 for(int i=1;i<=3;i++){ 91 for(int j=1;j<=3;j++){ 92 if(i != j){ 93 if(ans >= mid.a[i][j]){ 94 ans = mid.a[i][j]; 95 ins = i; 96 } 97 } 98 } 99 } 100 dfs2(be, 0, ins, ans, -1); 101 printf("%lld\n", ans); 102 for(int i=1;i<=n;i++)printf("%d%c", av[i], i==n ? '\n' : ' '); 103 104 return 0; 105 }
看这个有序列里两个最值的数量，以此更新更小长度的最值即可，利用multiset来写，代码简洁很多
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e5+7; 5 const ll mod = 1e9 + 9; 6 #define afdafafafdafaf y1; 7 int ar[maxn], n; 8 9 ll k; 10 int main() 11 { 12 scanf("%d%lld", &n, &k); 13 multiset<ll> s; 14 for(int i = 1; i <= n; i++){ 15 scanf("%d", ar + i); 16 s.insert(ar[i]); 17 } 18 ll be = * s.begin(); 19 auto it = s.end(); 20 it--; 21 ll en = *it; 22 ll be_ins = s.count(be), en_ins = s.count(en); 23 ll ans = 0; 24 s.erase(be); 25 s.erase(en); 26 while(be < en){ 27 if(s.size() == 0){ 28 be = min(en, be + k / min(be_ins, en_ins)); 29 break; 30 } 31 it = s.begin(); 32 ll new_be = *it; 33 it = s.end(); 34 it--; 35 ll new_en = *it; 36 if(be_ins < en_ins){ 37 if(k > be_ins * (new_be - be)){ 38 k -= be_ins * (new_be - be); 39 be = new_be; 40 be_ins += s.count(be); 41 s.erase(be); 42 } 43 else{ 44 be = min(en, be + k / be_ins); 45 break; 46 } 47 } 48 else{ 49 if(k > en_ins * (en - new_en)){ 50 k -= en_ins * (en - new_en); 51 en = new_en; 52 en_ins += s.count(en); 53 s.erase(en); 54 } 55 else{ 56 en = max(be, en - k / en_ins); 57 break; 58 } 59 } 60 } 61 ans = en - be; 62 printf("%lld\n", max(0LL, ans)); 63 return 0; 64 }
