Day4 morning Problem Solving report, day4 morning Problem Solving

Day4 morning Problem Solving report, day4 morning Problem Solving
Expected score: 50 + 0 + 0 = 50 actual score: 50 + 0 + 10 = 60 subjects with cancer, T3 does not give brute force score (* ￣) T1

Https://www.luogu.org/problem/show? Pid = T15564

Greedy,

But I don't know how to maintain it.

 

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 # include <algorithm> 6 using namespace std; 7 const int MAXN = 1e6; 8 const int INF = 0x7ffff; 9 inline int read () 10 {11 char c = getchar (); int flag = 1, x = 0; 12 while (c <'0' | c> '9') {if (c = '-') flag =-1; c = getchar ();} 13 while (c> = '0' & c <= '9') x = x * 10 + c-48, c = getchar (); return x * flag; 14} 15 int zero; 16 int n; 17 int a [MAXN]; 18 int tmp [MAX N]; 19 int main () 20 {21 // freopen ("multiset. in "," r ", stdin); 22 // freopen (" multiset. out "," w ", stdout); 23 n = read (); 24 for (int I = 1; I <= n; I ++) a [I] = read (); 25 sort (a + 1, a + n + 1); 26 int now = 0; 27 int num = n; 28 int bg = INF; 29 int ans = 0; 30 while (num! = 1) 31 {32 zero = 0; bg = INF; 33 for (int I = 1; I <= num; I ++) 34 {35 if (a [I] = 0) zero ++; 36 if (a [I]> = 1) 37 {38 a [I] --; 39 if (bg = INF) bg = I; 40} 41 42} 43 int nownum = 0; 44 for (int I = 1; I <= zero/2; I ++) 45 tmp [++ nownum] = 0; 46 if (zero % 2 = 1) 47 tmp [++ nownum] = 0; 48 for (int I = bg; I <= num; I ++) 49 tmp [++ nownum] = a [I]; 50 for (int I = 1; I <= nownum; I ++) a [I] = tmp [I]; 51 num = nownum; 52 ans ++; 53} 54 printf ("% d ", ans); 55 return 0; 56} 57 58 59/* 60 561 0 0 3 3 // 562 263 2 1 // 364 465 1 0 2 3 // 466 67 */50 points of violence

 

Positive Solution

Consider how to move from the final state

Greedy ideas:

Make the value not 0 smaller.

Reduce the number of zeros.

Violence-"50

How to optimize

Use a bucket to record,

A [0]-> (a [0] + 1)/2-> a [1]

 

1 # include <cstdio> 2 # include <iostream> 3 # include <algorithm> 4 # include <cstring> 5 # include <climits> 6 # include <cstdlib> 7 # include <cmath> 8 # include <queue> 9 # include <vector> 10 # include <utility> 11 using namespace std; 12 const int MAXN = 5e6 + 60, MX = 1e6; 13 int N, a [MAXN], cnt [MAXN], res, lim; 14 15 int main () 16 {17 freopen ("multiset. in "," r ", stdin); 18 freopen (" multiset. out "," w ", stdout); 19 scanf (" % d ", & N); 20 for (int I = 1; I <= N; ++ I) {21 scanf ("% d", & a [I]); 22 lim = max (lim, a [I]); 23 ++ cnt [a [I]; 24} 25 int l = 0, z = cnt [0]; 26 for (int I = 1; I <= lim; ++ I) {27 ++ res; 28 z = (z + 1)/2; 29 z + = cnt [I]; 30} 31 for (; z> 1; z = (z + 1)/2) + res; 32 printf ("% d \ n", res); 33 return 0; 34}View Code

 

T2

 

I didn't understand the example at the beginning.

For a long time, I thought he meant to re-mark each group ..

It wasn't until I asked zzx in the last 20 minutes that I found that there was a brute force score of 50 !!!! (Lyq has been tested for 80) {{{( >_< )}}}

 

Positive Solution:

Method 1: Add edge in brute force mode. Each BFS operation can add a pruning rule. BFS is used only for vertices that need to be added each time. Complexity: $ O (m) $

Method 2: one-way parallel query

Method 3: For each vertex

Multiply the length of the split to ensure that the length is between $2 ^ ~ 2 ^ p $, second in this interval, time complexity $ O (m * log) $

 

1 # include <iostream> 2 # include <cstdio> 3 # include <cstdlib> 4 # include <cstring> 5 # include <algorithm> 6 # include <cmath> 7 # include <vector> 8 using namespace std; 9 const int N = 500010; 10 typedef long ll; 11 12 int n, m; 13 int vis [N], u [N], v [N]; 14 vector <int> vec [N]; 15 16 bool dfs (int u) 17 {18 if (vis [u]) return false; 19 if (u = n) return true; 20 vis [u] = 1; 21 bool ret = false; 22 for (int I = 0; I <vec [u]. size (); I ++) 23 {24 ret = dfs (vec [u] [I]); 25 if (ret) return true; 26} 27 return false; 28} 29 30 int read () 31 {32 char ch = getchar (); 33 int x = 0; 34 while (! Isdigit (ch) ch = getchar (); 35 while (isdigit (ch) {x = x * 10 + (ch-'0 '); ch = getchar () ;}36 return x; 37} 38 39 bool check (int sta, int las) 40 {41 for (int I = sta; I <= las; I ++) 42 vec [u [I]. push_back (v [I]); 43 44 bool ret = dfs (1); 45 46 for (int I = sta; I <= las; I ++) 47 {48 vis [u [I] = vis [v [I] = 0; 49 vec [u [I]. clear (); 50} 51 vis [1] = 0; 52 return ret; 53} 54 55 56 57 int main () 58 {59 freopen ("road. in ", "R", stdin); 60 freopen ("road. out "," w ", stdout); 61 n = read (), m = read (); 62 for (int I = 0; I <m; I ++) 63 {64 // scanf ("% d", & u [I], & v [I]); 65 u [I] = read (); v [I] = read (); 66} 67 int now = 0, ans = 0; 68 while (now <m) 69 {70 int I; 71 for (I = 1; I + now <= m; I <= 1) 72 if (check (now, now + I-1) break; 73 I> = 1; 74 int nowtmp = now + I; 75 for (; I> 0; I> = 1) 76 if (nowtmp + I <= m &&! Check (now, nowtmp + I-1) 77 nowtmp + = I; 78 ans ++; 79 now = nowtmp; 80 81} 82 cout <ans <endl; 83}Positive Solution

 

 

 

T3

At first, yy had a self-righteous greed.

Then it took more than an hour to get violent.

Only n <= 10 can be detected. Greedy groups have two wrong groups.

GG...

Later, it was inexplicably more than a point ....

 

 

Positive Solution

Greedy

Expected sequence: 1 2 3 4 5

5 5 5 5 5

4 3 2 1 0

1 2 3 4 5

$ F [I] [j] $ a soldier with a blood volume of $ I $, empty $ j $ knife

Deformation of the backpack ..

Maintain with one stack

 

1 # include <iostream> 2 # include <cstdio> 3 # include <cstdlib> 4 # include <algorithm> 5 # include <cmath> 6 # include <cstring> 7 using namespace std; 8 9 const int MAXN = 1000 + 10; 10 int a [MAXN]; 11 int cnt [MAXN], sta [MAXN], c [MAXN]; 12 int f [MAXN] [MAXN]; 13 14 int main () 15 {16 freopen ("cs. in "," r ", stdin); 17 freopen (" cs. out "," w ", stdout); 18 int T; 19 scanf (" % d ", & T); 20 for (int num = 1; num <= T; ++ n Um) 21 {22 int N, maxn = 0; 23 scanf ("% d", & N); 24 memset (cnt, 0, sizeof (cnt )); 25 memset (c, 0, sizeof (c); 26 memset (f, 0, sizeof (f); 27 for (int I = 1; I <= N; ++ I) 28 {29 scanf ("% d", & a [I]); ++ cnt [a [I]; 30 maxn = max (maxn, a [I]); 31} 32 int top = 0; 33 for (int I = 1; I <= maxn; ++ I) 34 if (cnt [I] = 0) sta [++ top] = I; else35 {36 while (cnt [I]> 1 & top> 0) {c [sta [top --] = I; -- cnt [I];} 37 c [I] = I; 38} 39 40 int ans = 0; 41 for (int I = 1; I <= maxn; ++ I) 42 for (int j = 0; j <= I; ++ j) 43 {44 if (j> 0) f [I] [j] = f [I-1] [j-1]; 45 if (c [I]! = 0 & j + c [I]-I <I) 46 f [I] [j] = max (f [I] [j], f [I-1] [j + c [I]-I] + 1); 47 ans = max (ans, f [I] [j]); 48} 49 printf ("% d \ n", ans); 50} 51 return 0; 52}View Code

 

 

 

Summary

This exam made too many mistakes ..

T2 clearly has 80 brute force points

It's about T3 ....