POI #Year2006 #贪心 #暴力
如果将所有的列干掉,那么贪心的如果可以删除一列就优先删除一列
当左右列不可以删除时,考虑上下行先删除谁不好判断
考虑在操作之前先枚举上面删除多少行,这样在操作时就可以贪心的使得下面删掉的行尽可能少
也就是贪心先删除上面
// Author: xiaruize
#ifndef ONLINE_JUDGE
#define debug(...) std::cerr << __LINE__ << ": [", __DEBUG_UTIL__::printer(#__VA_ARGS__, __VA_ARGS__)
#define debugArr(...) std::cerr << __LINE__ << ": [", __DEBUG_UTIL__::printerArr(#__VA_ARGS__, __VA_ARGS__)
#else
#define debug(...)
#define debugArr(...)
#endif
using namespace std;
#ifndef ONLINE_JUDGE
clock_t start_clock = clock();
#endif
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
if (a > b)
return a;
return b;
}
int min(int a, int b)
{
if (a < b)
return a;
return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e3 + 10;
int k, n, m;
int a[N][N], b[N][N];
int rw[N][N], col[N][N];
int calc1(int x)
{
int u = 1, d = n;
int res = 0;
for (int l = 1, r = m; l <= r;)
{
while (l <= r && col[d][l] - col[u - 1][l] <= k)
l++;
while (l <= r && col[d][r] - col[u - 1][r] <= k)
r--;
// debug(x, l, r, sum[d][r] - sum[u - 1][r]);
if (l > r)
break;
if (u < x && rw[u][r] - rw[u][l - 1] <= k)
{
u++;
res++;
}
else if (u <= d && rw[d][r] - rw[d][l - 1] <= k)
{
res++;
d--;
}
else
return INF;
}
return res + m;
}
void solve()
{
cin >> k >> m >> n;
rep(i, 1, n)
{
rep(j, 1, m)
{
cin >> a[i][j];
b[j][i] = a[i][j];
rw[i][j] = a[i][j] + rw[i][j - 1];
col[i][j] = a[i][j] + col[i - 1][j];
// debug(sum[i][j]);
}
}
int res = INF;
rep(i, 1, n) res = min(res, calc1(i));
// cerr << res << endl;
swap(n, m);
mms(a, 0);
mms(rw, 0);
mms(col, 0);
rep(i, 1, n)
{
rep(j, 1, m)
{
a[i][j] = b[i][j];
rw[i][j] = a[i][j] + rw[i][j - 1];
col[i][j] = a[i][j] + col[i - 1][j];
// debug(sum[i][j]);
}
}
rep(i, 1, n) res = min(res, calc1(i));
cout << res << endl;
}
#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif
signed main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int testcase = 1;
// cin >> testcase;
while (testcase--)
solve();
#ifndef ONLINE_JUDGE
cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
return 0;
}
标签:rw,return,删除,Ploughing,int,res,POI2006ORK,define
From: https://www.cnblogs.com/xiaruize/p/18136789