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  • 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 '
    '
     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
    ", 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
    ", 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ii个里面经历$k$次出逃可以取到最少的修改数
     
    转移什么的就太简单了不说了自己看代码吧
      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 '
    '
     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
    ", 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
    ", 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 '
    '
     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
    ", 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
    ", 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

    加油ヾ(◍°∇°◍)ノ゙
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  • 原文地址:https://www.cnblogs.com/harryhqg/p/10038392.html
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