呐呐呐~第一份随笔
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算法被发明出来，是为了解决问题；而评价一个算法的好坏，在于它能否迅速、简洁、优雅地解决问题。
“保持一颗热爱的心。你也能够成功的吧。”
这是学习算法的初衷。相信自己吧。
PART 1 预备知识
<1>
我们常常需要（1）快读（2）重定向（3）打表（4）预处理 等等，这些都是对数据读写的处理。
（1）快读常用于读入数据量较多的 int, 它的原理由 getchar() 实现；
inline int read()
{
int s = 1, v = 0;
char c = getchar();
while (c < '0' || c > '9')
{
if (c == '-')
s = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
v = (v << 3) + (v << 1) + (c ^ 48);
c = getchar();
}
return s * v;
}
（2）重定向常用于文件读写，它通过 freopen() 实现；
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
（3）打表用于解决时间复杂度高且数据范围小的问题，将所有结果输出并记录在数组里；
（4）预处理是将一些可能会用到的数据提前处理并记录下来，便于以后调用，如埃氏筛、欧拉筛。
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
int n, cnt, a[N], p[N];
void Era()
{
for (int i = 2; i <= N; ++i)
{
if (!a[i])
p[cnt++] = i;
for (int j = i; j <= N; j += i)
a[j] = 1;
}
}
int main()
{
cin >> n;
Era();
cout << p[n];
}
点击查看代码
#include <bits/stdc++.h>
using namespace std;
const int N = 1e8;
int n, cnt, a[N], p[N];
void Euler()
{
for (int i = 2; i <= N; ++i)
{
if (!a[i])
p[cnt++] = i;
for (int j = 0; j < cnt; j++)
{
if (i * p[j] > N)
break;
a[i * p[j]] = true;
if (i % p[j] == 0)
break;
}
}
}
int main()
{
cin >> n;
Euler();
cout << p[n];
}
<2>
一些经典问题的解法是值得背诵的。
（1）八皇后（深度优先搜索，状态压缩)
#include <bits/stdc++.h>
using namespace std;
int N, cnt, b1[100], b2[100], b3[100];
void dfs(int n)
{
if (n > N)
{
cnt++;
return;
}
for (int i = 1; i <= N; i++)
if (b1[i] == 0 && b2[N - n + i] == 0 && b3[n + i] == 0)
{
b1[i] = 1, b2[N - n + i] = 1, b3[n + i] = 1;
dfs(n + 1);
b1[i] = 0, b2[N - n + i] = 0, b3[n + i] = 0;
}
}
int main()
{
cin >> N;
dfs(1);
cout << cnt;
}
点击查看代码
#include <bits/stdc++.h>
using namespace std;
unsigned long long res, k, n, cnt, num[1000], s[1000], f[10][300][300];
void init()
{
for (int i = 0; i < (1 << n); i++)
{
if (i & (i << 1))
continue;
int sum = 0;
for (int j = 1; j < (1 << n); j <<= 1)
{
if (j & i)
sum++;
}
num[++cnt] = sum;
s[cnt] = i;
f[1][cnt][sum] = 1;
}
}
void dp()
{
for (int i = 1; i <= n; i++)
for (int j = 1; j <= cnt; j++)
for (int r = 0; r <= k; r++)
if (r >= num[j])
for (int q = 1; q <= cnt; q++)
if (!(s[j] & s[q]) && !((s[j] << 1) & s[q]) && !((s[j] >> 1) & s[q]))
f[i][j][r] += f[i - 1][q][r - num[j]];
for (int i = 1; i <= cnt; i++)
res += f[n][i][k];
}
int main()
{
cin >> n >> k;
init();
dp();
cout << res;
}
（2）迷宫（广度优先搜索）
#include <bits/stdc++.h>
using namespace std;
int n, m, sx, sy, ex, ey;
int a[100][100], b[100][100], d[4][2] = {{1, 0}, {0, -1}, {-1, 0}, {0, 1}};
struct node
{
int x, y, step;
};
queue<node> q;
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
cin >> a[i][j];
cin >> sx >> sy >> ex >> ey;
q.push({sx, sy, 0});
while (!q.empty())
{
int x = q.front().x, y = q.front().y, step = q.front().step;
q.pop();
if (x == ex && y == ey)
{
cout << step;
return 0;
}
for (int i = 0; i < 4; i++)
{
int tx = x + d[i][0], ty = y + d[i][1];
if (a[tx][ty] == 1 && b[tx][ty] == 0)
{
b[tx][ty] = 1;
q.push({tx, ty, step + 1});
}
}
}
cout << "-1";
}
（3）路标设置（二分法）
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
int l, n, k, ans, a[N];
bool check(int x)
{
int i = 1, cnt = 0;
while (i < n)
{
int j = 1;
while (a[i] - a[i - 1] > j * x)
{
cnt++;
j++;
}
i++;
}
if (cnt <= k)
return true;
else
return false;
}
int main()
{
cin >> l >> n >> k;
for (int i = 0; i < n; i++)
cin>>a[i];
int left = 1, right = l;
while (left <= right)
{
int mid = (left + right) / 2;
if (check(mid))
{
right = mid - 1;
ans = mid;
}
else
left = mid + 1;
}
cout << ans;
}
（4）背包（记忆化搜索，动态规划）
#include <bits/stdc++.h>
using namespace std;
int n, W, w[100], v[100], f[1000][1000];
int dfs(int i, int j)
{
if (f[i][j] >= 0)
return f[i][j];
int res;
if (i == n)
res = 0;
else if (j < w[i])
res = dfs(i + 1, j);
else
res = max(dfs(i + 1, j), dfs(i + 1, j - w[i]) + v[i]);
return f[i][j] = res;
}
int main()
{
cin >> n;
for (int i = 0; i < n; i++)
cin >> w[i] >> v[i];
cin >> W;
memset(f, -1, sizeof(f));
cout << dfs(0, W);
}
点击查看代码
#include <bits/stdc++.h>
using namespace std;
int n, W, w[100], v[100], f[1000][1000];
int main()
{
cin >> n;
for (int i = 0; i < n; i++)
cin >> w[i] >> v[i];
cin >> W;
for (int i = n - 1; i >= 0; i--)
for (int j = 0; j <= W; j++)
if (j < w[i])
f[i][j] = f[i + 1][j];
else
f[i][j] = max(f[i + 1][j], f[i + 1][j - w[i]] + v[i]);
cout << f[0][W];
}
（5）最小逆序对数（分治）
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e6 + 5;
int t, n, m, tot, cnt;
int a[N], b[N], c[N], d[N], A[N], p[N];
void disc()
{
sort(d + 1, d + 1 + tot);
cnt = unique(d + 1, d + 1 + tot) - d - 1;
for (int i = 1; i <= n; i++)
a[i] = lower_bound(d + 1, d + 1 + cnt, a[i]) - d;
for (int i = 1; i <= m; i++)
b[i] = lower_bound(d + 1, d + 1 + cnt, b[i]) - d;
}
int lowbit(int x)
{
return x & (-x);
}
void upd(int x)
{
for (; x <= cnt; x += lowbit(x))
c[x]++;
}
ll qry(int x)
{
ll summ = 0;
for (; x; x -= lowbit(x))
summ += c[x];
return summ;
}
ll calc()
{
ll summ = 0;
for (int i = 1; i <= tot; i++)
{
summ += qry(cnt) - qry(A[i]);
upd(A[i]);
}
return summ;
}
void solve(int Lb, int Rb, int La, int Ra)
{
if (Lb > Rb)
return;
int mid = (Lb + Rb) >> 1;
int val = b[mid];
int minn = 2e9, summ1 = 0, summ2 = 0;
for (int i = Ra - 1; i >= La; i--)
{
if (a[i] < val)
summ2++;
}
minn = summ2;
p[mid] = La;
for (int i = La + 1; i <= Ra; i++)
{
if (a[i - 1] < val)
summ2--;
if (a[i - 1] > val)
summ1++;
if (summ1 + summ2 < minn)
{
minn = summ1 + summ2;
p[mid] = i;
}
}
solve(Lb, mid - 1, La, p[mid]);
solve(mid + 1, Rb, p[mid], Ra);
}
int main()
{
cin >> t;
while (t--)
{
tot = 0;
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
d[++tot] = a[i];
}
for (int i = 1; i <= m; i++)
{
cin >> b[i];
d[++tot] = b[i];
}
sort(b + 1, b + 1 + m);
disc();
for (int i = 1; i <= cnt; i++)
c[i] = 0;
solve(1, m, 1, n + 1);
tot = 0;
int k = 1;
for (int i = 1; i <= n; i++)
{
while (p[k] == i && k <= m)
{
A[++tot] = b[k];
k++;
}
A[++tot] = a[i];
}
while (k <= m)
{
A[++tot] = b[k];
k++;
}
cout << calc();
}
}
标签：呐呐,include,int,cin,++,第一份,cnt,100,随笔 来源： https://www.cnblogs.com/Arcticus/p/16066933.html