链接:HDU6794 Tokitsukaze and Multiple
#include
using namespace std;
const int MAXN = 1e5 + 7;
int T;
int n, p;
int arr[MAXN];
bool vis[MAXN];
int mod[MAXN];
int ans;
int main(void)
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> T;
while (T--)
{
ans = 0;
memset(vis, 0, sizeof(vis));
memset(mod, 0, sizeof(mod));
cin >> n >> p;
for (int i = 1; i <= n; ++i)
{
cin >> arr[i];
arr[i] %= p;
if (!arr[i])
{
vis[i] = true;
ans++;
}
}
int s = 0; // Start of the non-visited Interval
int sum = 0; // Sum of the non-visited Interval
for (int i = 1; i <= n; ++i)
{
if (!vis[i])
{
sum += arr[i];
sum %= p;
if (sum == 0 || mod[sum % p] > s)
{
ans++;
vis[i] = true;
s = i;
sum = 0;
continue;
}
mod[sum % p] = i;
}
if (vis[i])
{
s = i;
sum = 0;
}
}
cout << ans << endl;
}
return 0;
}
链接:HDU6795 Little W and Contest
因为负数取模的原因WA了5发
#include
using namespace std;
#define int long long
const int MAXN = 1e5 + 7;
const int MOD = 1e9 + 7;
int fac[MAXN], inv[MAXN];
inline void Pre(int n)
{
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = fac[i - 1] * i % MOD;
inv[1] = 1;
for (int i = 2; i <= n; i++)
inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
inv[0] = 1;
for (int i = 1; i <= n; i++)
inv[i] = inv[i] * inv[i - 1] % MOD;
}
inline int C(int n, int m)
{
if (m > n)
return 0;
return fac[n] * inv[m] % MOD * inv[n - m] % MOD;
}
// 计算组合数
inline int Pow(int a, int b)
{
int ret = 1;
for (; b; b >>= 1, a = a * a % MOD)
if (b & 1)
ret = ret * a % MOD;
return ret;
}
// 快速幂
int T;
int n;
int pts[MAXN];
int u, v;
int ans;
int tmp1, tmp2;
struct DSU
{
int parent[MAXN];
int n; // Num of Nodes
int cnt; // Count Strongly Connected Components
int sum1[MAXN];
int sum2[MAXN];
int rank[MAXN];
void init_parent(int n)
{
this -> n = n;
cnt = 0;
for (int i = 1; i <= n; ++i)
{
parent[i] = i;
if (pts[i] == 1)
{
sum1[i] = 1;
sum2[i] = 0;
}
else
{
sum1[i] = 0;
sum2[i] = 1;
}
rank[i] = 0;
}
}
int find_parent(int x)
{
if (parent[x] == x)
return x;
else
parent[x] = find_parent(parent[x]);
return parent[x];
}
bool check_unicom(int x, int y)
{
return find_parent(x) == find_parent(y);
}
void merge(int x, int y)
{
x = find_parent(x);
y = find_parent(y);
if (rank[x] > rank[y])
{
parent[y] = x;
if (x != y)
{
sum1[x] += sum1[y];
sum1[x] %= MOD;
sum2[x] += sum2[y];
sum2[x] %= MOD;
}
}
else
{
parent[x] = y;
if (x != y)
{
sum1[y] += sum1[x];
sum1[y] %= MOD;
sum2[y] += sum2[x];
sum2[y] %= MOD;
}
if (rank[x] == rank[y])
rank[y] += 1;
}
}
int getsum1(int i)
{
return sum1[find_parent(i)];
}
int getsum2(int i)
{
return sum2[find_parent(i)];
}
}dsu;
int32_t main(void)
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
Pre(100005);
cin >> T;
while (T--)
{
cin >> n;
ans = 0;
tmp1 = tmp2 = 0;
for (int i = 1; i <= n; ++i)
{
cin >> pts[i];
if (pts[i] == 1)
{
tmp1++;
}
if (pts[i] == 2)
{
tmp2++;
}
}
dsu.init_parent(n);
ans = (C(tmp2, 2) % MOD * tmp1 % MOD) % MOD + C(tmp2, 3) % MOD;
ans %= MOD;
cout << ans << endl;
for (int i = 1; i <= n - 1; ++i)
{
cin >> u >> v;
if (!dsu.check_unicom(u, v))
{
int usum1 = dsu.getsum1(u);
int usum2 = dsu.getsum2(u);
int vsum1 = dsu.getsum1(v);
int vsum2 = dsu.getsum2(v);
// cout << usum1 << " " << usum2 << " " << vsum1 << " " << vsum2 << endl;
int les2 = tmp2 - usum2 - vsum2;
int les1 = tmp1 - usum1 - vsum1;
// cout << les1 << " " << les2 << endl;
ans = ans + 2 * MOD
// Incorrect Pos
- les2 * ((usum2 * vsum1) % MOD + (usum1 * vsum2) % MOD) % MOD
- ((les2 + les1) % MOD) * (usum2 * vsum2 % MOD) % MOD
;
dsu.merge(u, v);
}
ans %= MOD;
cout << ans << endl;
}
}
return 0;
}
链接:HDU6799 Parentheses Matching
*
(
)
元素*
可以替换成任意一个括号(
从右侧加)
(队友的代码)
#include
#include
#include
using namespace std;
stack<char> st;
char ans[100007];
int main()
{
int t;
string s, ans;
cin >> t;
while (t--)
{
cin >> s;
while (!st.empty())
st.pop();
int cnt1 = 0, cnt2 = 0;
bool flag = true;
for (int i = 0; s[i]; i++)
{
if (s[i] == '*')
cnt1++;
else if (s[i] == '(')
st.push(s[i]);
else
{
if (!st.empty())
st.pop();
else
{
cnt2++;
if (cnt2 > cnt1)
{
flag = false;
break;
}
}
}
}
if (!flag)
{
printf("No solution!\n");
continue;
}
else
{
int cnt = 0;
for (int i = 0; s[i] && cnt < cnt2; i++)
{
if (s[i] == '*')
{
cnt++;
s[i] = '(';
}
}
while (!st.empty())
st.pop();
cnt = 0;
cnt1 = 0;
cnt2 = 0;
int len = s.length();
for (int i = len - 1; i >= 0; i--)
{
if (s[i] == ')')
{
st.push(s[i]);
}
else if (s[i] == '*')
cnt1++;
else
{
if (!st.empty())
st.pop();
else
{
cnt2++;
if (cnt2 > cnt1)
{
flag = false;
break;
}
}
}
}
// cout << s << endl;
if (!flag)
{
printf("No solution!\n");
continue;
}
for (int i = len - 1; i >= 0 && cnt < cnt2; i--)
{
if (s[i] == '*')
{
cnt++;
s[i] = ')';
}
}
// cout << s << endl;
for (int i = 0; s[i]; i++)
{
if (s[i] != '*')
printf("%c", s[i]);
}
cout << endl;
}
}
return 0;
}
链接:HDU6797 Tokitsukaze and Rescue
n n n 个点,每个点间均有一条无向边 ( n ⋅ ( n − 1 ) / 2 n \cdot (n - 1) / 2 n⋅(n−1)/2 条无向边)所构成的无向图
求删去任意 m m m 条边后, 1 − n 1 - n 1−n 最短路的最大距离
每条边权随机 → \to → 边权随机的情况下,最短路的边数很少。
(这个怎么证啊)
#include
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 50 + 7;
const int MAXE = 1E5 + 7;
int ans;
struct Dijkstra
{
int graph[MAXN][MAXN]; // Adj Matrix
int n; // Num of Nodes
int s, t; // Start Pos and End Pos
bool vis[MAXN];
int dis[MAXN];
int path[7][MAXN];
void init()
{
memset(graph, 0x3f, sizeof(graph));
memset(path, -1, sizeof(path));
for (int i = 1; i <= n; ++i)
{
graph[i][i] = 0;
}
s = 1;
t = n;
}
void dijkstra(int m)
{
memset(vis, 0, sizeof(vis));
memset(dis, 0x3f, sizeof(dis));
vis[s] = true;
dis[s] = 0;
for (int i = 1; i <= n; ++i)
{
if (dis[i] > dis[s] + graph[s][i])
{
dis[i] = dis[s] + graph[s][i];
path[m][i] = s;
}
}
while (true)
{
int nxt = -1;
for (int i = 1; i <= n; ++i)
{
if (!vis[i] && (nxt == -1 || dis[nxt] > dis[i]))
{
nxt = i;
}
}
if (nxt == -1)
{
break;
}
vis[nxt] = true;
for (int i = 1; i <= n; ++i)
{
if (!vis[i] && dis[i] > dis[nxt] + graph[nxt][i])
{
dis[i] = dis[nxt] + graph[nxt][i];
path[m][i] = nxt;
}
}
}
}
void dfs(int m)
{
dijkstra(m);
if (m == 0)
{
ans = max(ans, dis[t]);
return;
}
for (int i = t; path[m][i] != -1; i = path[m][i])
{
int fr = path[m][i];
int tmp = graph[fr][i];
graph[fr][i] = INF;
graph[i][fr] = INF;
dfs(m - 1);
graph[fr][i] = tmp;
graph[i][fr] = tmp;
}
}
}djk;
int T;
int n, m;
int u, v, val;
int main(void)
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> T;
while (T--)
{
cin >> n >> m;
djk.n = n;
djk.init();
ans = 0;
for (int i = 1; i <= n * (n - 1) / 2; i++)
{
cin >> u >> v >> val;
djk.graph[u][v] = djk.graph[v][u] = val;
}
djk.dfs(m);
cout << ans << endl;
}
return 0;
}
链接:HDU6798 Triangle Collision
(待补)