USACO 2018 February Contest, Gold 总结

这一场对我来说要比上一场稍难一点，

P1 Snow Boots

还是一如既往地简单，有些人用什么线段树，整体二分，单调队列，你们脑子坏了么？

为什么不用并查集呢！！！

先都按S排序

每次把不能走的染黑，如果任意一段连续的黑色个数小于< d的话我就可以通过，否则不行。

用并查集来维护最大子段和。

时间复杂度$O(nlogn)$

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i) = (a); (i) <= (b); (i)++)
 22 #define Rep(i, a, b) for (int (i) = (a) - 1; (i) < (b); (i)++)
 23 #define REP(i, a, b) for (int (i) = (a); (i) >= (b); (i)--)
 24 #define clr(a) memset(a, 0, sizeof(a))
 25 #define Sort(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 26 #define ass(a, sum) memset(a, sum, sizeof(a))
 27 
 28 #define ls ((rt) << 1)
 29 #define rs ((rt) << 1 | 1)
 30 #define lowbit(x) (x & -x)
 31 #define mp make_pair
 32 #define pb push_back
 33 #define fi first
 34 #define se second
 35 #define endl '\n'
 36 #define ENDL cout << endl
 37 #define SZ(x) ((int)x.size())
 38 
 39 typedef long long ll ;
 40 typedef unsigned long long ull ;
 41 typedef vector <int> vi ;
 42 typedef pair <int, int> pii ;
 43 typedef pair <ll, ll> pll ;
 44 typedef map <int, int> mii ;
 45 typedef map <string, int> msi ;
 46 typedef map <ll, ll> mll ;
 47 
 48 const int N = 100010 ;
 49 const double eps = 1e-8 ;
 50 const int iinf = INT_MAX ;
 51 const ll linf = 2e18 ;
 52 const double dinf = 1e30 ;
 53 const int MOD = 1000000007 ;
 54 
 55 inline int read(){
 56     int X = 0, w = 0 ;
 57     char ch = 0 ;
 58     while (!isdigit(ch)) { w |= ch == '-' ; ch = getchar() ; }
 59     while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar() ;
 60     return w ? - X : X ;
 61 }
 62 
 63 void write(int x){
 64      if (x < 0) putchar('-'), x = - x ;
 65      if (x > 9) write(x / 10) ;
 66      putchar(x % 10 + '0') ;
 67 }
 68 
 69 void print(int x) {
 70     cout << x << endl ;
 71     exit(0) ;
 72 }
 73 
 74 void PRINT(string x) {
 75     cout << x << endl ;
 76     exit(0) ;
 77 }
 78 
 79 void douout(double x){
 80      printf("%lf\n", x + 0.0000000001) ;
 81 }
 82 
 83 int n, m ;
 84 int fa[N], sz[N], vis[N], ans[N] ;
 85 
 86 struct roads {int d, id ;} a[N] ;
 87 struct boots {int d, s, id ;} b[N] ;
 88 bool cmp1(roads a, roads b) {return a.d > b.d ;}
 89 bool cmp2(boots a, boots b) {return a.d > b.d ;}
 90 
 91 int find(int x) {
 92     return fa[x] == x ? x : fa[x] = find(fa[x]) ;
 93 }
 94 
 95 void Merge(int x, int y) {
 96     sz[y] += sz[x] ; fa[x] = y ;
 97 }
 98 
 99 signed main(){
100     freopen("snowboots.in", "r", stdin) ;
101     freopen("snowboots.out", "w", stdout) ;
102     scanf("%d%d", &n, &m) ;
103     for (int i = 1; i <= n; i++) scanf("%d", &a[i].d), a[i].id = i ;
104     for (int i = 1; i <= m; i++) scanf("%d %d", &b[i].d, &b[i].s), b[i].id = i ;
105     sort(a + 1, a + n + 1, cmp1) ;
106     sort(b + 1, b + m + 1, cmp2) ;
107     for (int i = 1; i <= n; i++) fa[i] = i, sz[i] = 1 ;
108     int cnt = 1, dis = 0 ;
109     for (int i = 1; i <= m; i++) {
110         while (cnt <= n && a[cnt].d > b[i].d) {
111             int x = a[cnt].id ; vis[x] = 1 ;
112             if (vis[x - 1]) Merge(x - 1, x) ;
113             if (vis[x + 1]) Merge(x, find(x + 1)) ;
114             dis = max(dis, sz[find(x)]) ;
115             cnt++ ;
116         }
117         if (dis < b[i].s) ans[b[i].id] = 1 ;
118     }
119     for (int i = 1; i <= m; i++) printf("%d\n", ans[i]) ;
120 }
121 
122 /*
123 写代码时请注意：
124     1.是否要开Long Long？数组边界处理好了么？
125     2.实数精度有没有处理？
126     3.特殊情况处理好了么？
127     4.做一些总比不做好。
128 思考提醒：
129     1.最大值和最小值问题可不可以用二分答案？
130     2.有没有贪心策略？否则能不能dp？
131 */

AC CODE

P3 Taming the Herd

我看了一眼T2，感觉不会做，就看T3.这个T3一眼看感觉就是dp把，然后数据范围提示是$O(n^3)$

思考了10分钟，大概想出来了一个状态：

$f [i] [j]$表示在前

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i) = (a); (i) <= (b); (i)++)
 22 #define Rep(i, a, b) for (int (i) = (a) - 1; (i) < (b); (i)++)
 23 #define REP(i, a, b) for (int (i) = (a); (i) >= (b); (i)--)
 24 #define clr(a) memset(a, 0, sizeof(a))
 25 #define Sort(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 26 #define ass(a, sum) memset(a, sum, sizeof(a))
 27 
 28 #define ls ((rt) << 1)
 29 #define rs ((rt) << 1 | 1)
 30 #define lowbit(x) (x & -x)
 31 #define mp make_pair
 32 #define pb push_back
 33 #define fi first
 34 #define se second
 35 #define endl '\n'
 36 #define ENDL cout << endl
 37 #define SZ(x) ((int)x.size())
 38 
 39 typedef long long ll ;
 40 typedef unsigned long long ull ;
 41 typedef vector <int> vi ;
 42 typedef pair <int, int> pii ;
 43 typedef pair <ll, ll> pll ;
 44 typedef map <int, int> mii ;
 45 typedef map <string, int> msi ;
 46 typedef map <ll, ll> mll ;
 47 
 48 const int N = 110 ;
 49 const double eps = 1e-8 ;
 50 const int iinf = INT_MAX ;
 51 const ll linf = 2e18 ;
 52 const double dinf = 1e30 ;
 53 const int MOD = 1000000007 ;
 54 
 55 inline int read(){
 56     int X = 0, w = 0 ;
 57     char ch = 0 ;
 58     while (!isdigit(ch)) { w |= ch == '-' ; ch = getchar() ; }
 59     while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar() ;
 60     return w ? - X : X ;
 61 }
 62 
 63 void write(int x){
 64      if (x < 0) putchar('-'), x = - x ;
 65      if (x > 9) write(x / 10) ;
 66      putchar(x % 10 + '0') ;
 67 }
 68 
 69 void print(int x) {
 70     cout << x << endl ;
 71     exit(0) ;
 72 }
 73 
 74 void PRINT(string x) {
 75     cout << x << endl ;
 76     exit(0) ;
 77 }
 78 
 79 void douout(double x){
 80      printf("%lf\n", x + 0.0000000001) ;
 81 }
 82 
 83 int n ;
 84 int a[N], b[N][N], dp[N][N] ;
 85 
 86 signed main(){
 87     freopen("taming.in", "r", stdin) ;
 88     freopen("taming.out", "w", stdout) ;
 89     scanf("%d", &n) ;
 90     for (int i = 1; i <= n; i++) scanf("%d", &a[i]) ;
 91     for (int i = 0; i <= n; i++) {
 92         int cnt = 0 ;
 93         for (int j = i; j <= n; j++) cnt += (a[j] != j - i), b[i][j] = cnt ;
 94     }
 95     ass(dp, 0x3f) ;
 96     dp[0][0] = 0 ;
 97     for (int i = 0; i <= n; i++)
 98     for (int j = 1; j <= n; j++)
 99     for (int k = i + 1; k <= n; k++)
100     dp[k][j] = min(dp[k][j], dp[i][j - 1] + b[i + 1][k]) ;
101     for (int i = 1; i <= n; i++) printf("%d\n", dp[n][i]) ;
102 }
103 
104 /*
105 写代码时请注意：
106     1.是否要开Long Long？数组边界处理好了么？
107     2.实数精度有没有处理？
108     3.特殊情况处理好了么？
109     4.做一些总比不做好。
110 思考提醒：
111     1.最大值和最小值问题可不可以用二分答案？
112     2.有没有贪心策略？否则能不能dp？
113 */

AC CODE

P2 Directory Traversal

this problem is a little difficult

即使我第一眼就看出来他是棵树（路径文件夹什么的不就是树状存储的么）

然后开始考虑贪心，发现不行，下意识的想树形dp

明显状态为 f[i]表示以i为出发点的答案

于是开始推式子，emmmm还想不太好玩

大概想了20分钟，中间也想过到底能不能dp

最终我还是推出来啦！

f[x]=f[fat[x]]-(len[x]+1)*size[x]+3*(leave-size[x])

其中几个数组解释一下

fat[x] x的father

size[x] x的子树大小

leave 直观意思，叶子

len[x] : 文件夹长度

然后就是两遍dfs扫出答案

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i) = (a); (i) <= (b); (i)++)
 22 #define Rep(i, a, b) for (int (i) = (a) - 1; (i) < (b); (i)++)
 23 #define REP(i, a, b) for (int (i) = (a); (i) >= (b); (i)--)
 24 #define clr(a) memset(a, 0, sizeof(a))
 25 #define Sort(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 26 #define ass(a, sum) memset(a, sum, sizeof(a))
 27 
 28 #define ls ((rt) << 1)
 29 #define rs ((rt) << 1 | 1)
 30 #define lowbit(x) (x & -x)
 31 #define mp make_pair
 32 #define pb push_back
 33 #define fi first
 34 #define se second
 35 #define endl '\n'
 36 #define ENDL cout << endl
 37 #define SZ(x) ((int)x.size())
 38 
 39 typedef long long ll ;
 40 typedef unsigned long long ull ;
 41 typedef vector <int> vi ;
 42 typedef pair <int, int> pii ;
 43 typedef pair <ll, ll> pll ;
 44 typedef map <int, int> mii ;
 45 typedef map <string, int> msi ;
 46 typedef map <ll, ll> mll ;
 47 
 48 const int N = 100010 ;
 49 const double eps = 1e-8 ;
 50 const int iinf = INT_MAX ;
 51 const ll linf = 2e18 ;
 52 const double dinf = 1e30 ;
 53 const int MOD = 1000000007 ;
 54 
 55 inline int read(){
 56     int X = 0, w = 0 ;
 57     char ch = 0 ;
 58     while (!isdigit(ch)) { w |= ch == '-' ; ch = getchar() ; }
 59     while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar() ;
 60     return w ? - X : X ;
 61 }
 62 
 63 void write(int x){
 64      if (x < 0) putchar('-'), x = - x ;
 65      if (x > 9) write(x / 10) ;
 66      putchar(x % 10 + '0') ;
 67 }
 68 
 69 void print(int x) {
 70     cout << x << endl ;
 71     exit(0) ;
 72 }
 73 
 74 void PRINT(string x) {
 75     cout << x << endl ;
 76     exit(0) ;
 77 }
 78 
 79 void douout(double x){
 80      printf("%lf\n", x + 0.0000000001) ;
 81 }
 82 
 83 int n, leave ;
 84 ll ans, f[N], dis[N] ;
 85 int len[N], sz[N] ;
 86 vector <int> son[N] ;
 87 
 88 void dfs1(int rt) {
 89     for (int i = 0; i < SZ(son[rt]); i++) {
 90         int to = son[rt][i] ;
 91         dis[to] = dis[rt] + len[to] + 1 ;
 92         dfs1(to) ;
 93         sz[rt] += sz[to] ;
 94     }
 95     if (!son[rt].size()) {
 96         sz[rt] = 1 ;
 97         dis[rt]-- ;
 98         f[1] += dis[rt] ;
 99     }
100 }
101 
102 void dfs2(int rt) {
103     for (int i = 0; i < SZ(son[rt]); i++) {
104         int to = son[rt][i] ;
105         if (!SZ(son[to])) continue ;
106         f[to] = f[rt] - (len[to] + 1) * sz[to] + 3 * (leave - sz[to]) ;
107         ans = min(ans, f[to]) ;
108         dfs2(to) ;
109     }
110 }
111 
112 int main() {
113     scanf("%d", &n) ;
114     for (int i = 1; i <= n; i++) {
115         int m ;
116         char s[20] ;
117         scanf("%s", s + 1) ;
118         len[i] = strlen(s + 1) ;
119         scanf("%d", &m) ;
120         if (!m) leave++ ;
121         for (int j = 1; j <= m; j++) {
122             int x ;
123             scanf("%d", &x) ;
124             son[i].pb(x) ;
125         }
126     }
127     dfs1(1) ;
128     ans = f[1] ;
129     dfs2(1) ;
130     printf("%lld\n", ans) ;
131 }
132 
133 /*
134 写代码时请注意：
135     1.是否要开Long Long？数组边界处理好了么？
136     2.实数精度有没有处理？
137     3.特殊情况处理好了么？
138     4.做一些总比不做好。
139 思考提醒：
140     1.最大值和最小值问题可不可以用二分答案？
141     2.有没有贪心策略？否则能不能dp？
142 */

AC CODE

USACO的题目再练练，今天居然被第二题给打了，不爽(＃￣～￣＃)，

加油

posted @ 2018-11-29 15:02 harryhqg 阅读(310) 评论(0) 编辑收藏举报

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USACO 2018 February Contest, Gold 总结

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