Submission #5270

#	Time	Username	Problem	Language	Result	Execution time	Memory
5270		tncks0121	Orchard (NOI14_orchard)	C++98	16 / 25	124 ms	14368 KiB

orchard

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 
#include <memory.h> 
#include <math.h> 
#include <assert.h> 
#include <stack> 
#include <queue> 
#include <map> 
#include <set> 
#include <algorithm> 
#include <string> 
#include <functional> 
#include <vector> 
#include <deque> 
#include <utility> 
#include <bitset> 
#include <limits.h>  
#include <time.h>

using namespace std; 
typedef long long ll; 
typedef unsigned long long llu; 
typedef double lf;
typedef unsigned int uint;
typedef long double llf;
typedef pair<int, int> pii;

int N, M;

namespace single_row {
	const int M_ = 1000005;
	const int LEAF = 1048576;
	int A[M_], S[M_];
	int T[M_];

	ll solve() {
		for(int i = 1; i <= M; i++) S[i] = S[i-1] + A[i];
		T[M+1] = (int)1e9;
		for(int j = M; j > 0; j--) T[j] = min(T[j+1], j - 2*S[j]);

		ll ret = N*M;
		for(int i = 1; i <= M; i++) {
			ret = min(ret, (ll)S[M] - (i-1) + 2*S[i-1] + T[i]);
		}

		return ret;
	}
};

namespace double_rows {
	const int M_ = 100005;
	int A[2][M_], S[M_], T[M_];

	ll solve() {
		ll ret = N*M;

		for(int i = 0; i < 2; i++) for(int j = 1; j <= M; j++) scanf("%d", &A[i][j]);
		for(int k = 0; k < 2; k++) {
			int cnt = 0;
			for(int i = 1; i <= M; i++) single_row::A[i] = A[k][i];
			for(int i = 1; i <= M; i++) cnt += A[!k][i];
			ret = min(ret, single_row::solve() + cnt);
		}

		for(int i = 1; i <= M; i++) S[i] = S[i-1] + A[0][i] + A[1][i];
		T[M+1] = (int)1e9;
		for(int j = M; j > 0; j--) T[j] = min(T[j+1], 2*j - 2*S[j]);

		for(int i = 1; i <= M; i++) {
			ret = min(ret, (ll)S[M] - 2*(i-1) + 2*S[i-1] + T[i]);
		}

		return ret;
	}
};

int main() {
	ll res;

	scanf("%d%d", &N, &M); res = (ll)N*M;
	if(N == 1) {
		for(int i = 1; i <= M; i++) scanf("%d", single_row::A+i);
		res = single_row::solve();
	}else if(N == 2){
		res = double_rows::solve();
	}else {

	}

	printf("%lld\n", res);

	return 0;
}

• Subtask 1: 4 / 4
• Subtask 2: 4 / 4
• Subtask 3: 5 / 5
• Subtask 4: 3 / 3
• Subtask 5: 0 / 4
• Subtask 6: 0 / 5