yolandaw

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yolandaw
15天前
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int n, k, t;
long long gcd(long long a,long long b)
{
return b == 0 ? a:gcd(b, a % b);
}

int main()
{

cin>>t;

while (t --)
{
cin >> n >> k;
int m = pow(10, k);
cout<< (long long) n * m / gcd(n, m) << endl;
}

return 0;
}


yolandaw
15天前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

int t;

int main()
{
cin >> t;
while (t -- )
{
LL a, b, k;
cin >> a >> b >> k;
LL c = a - b;
LL e = k / 2;
k %= 2;
if (k)
cout << c * e + a << endl;
else
cout << c * e << endl;
}
return 0;
}


yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 510, INF = 0x3f3f3f3f;

int n, m;
int g[N][N];
int dist[N];
bool st[N];

int prim()
{
memset(dist, 0x3f, sizeof dist);

int res = 0;
for (int i = 0; i < n; i ++ )
{
int t = -1;
for (int j = 1; j <= n; j ++ )
if (!st[j] && (t == -1 || dist[t] > dist[j]))
t = j;

if (i && dist[t] == INF) return INF;
if (i) res += dist[t];

for (int j = 1; j <= n; j ++ ) dist[j] = min(dist[j], g[t][j]);

st[t] = true;
}

return res;
}

int main()
{
cin >> n >> m;

memset(g, 0x3f, sizeof g);

while (m -- )
{
int a, b, c;
cin >> a >> b >> c;
g[a][b] = g[b][a] = min(g[a][b], c);
}

int t = prim();

if (t == INF) puts("impossible");
else cout << t;

return 0;
}


yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 510, INF = 0x3f3f3f3f;

int n, m;
int g[N][N];
int dist[N];
bool st[N];

int prim()
{
memset(dist, 0x3f, sizeof dist);

int res = 0;
for (int i = 0; i < n; i ++ )
{
int t = -1;
for (int j = 1; j <= n; j ++ )
if (!st[j] && (t == -1 || dist[t] > dist[j]))
t = j;

if (i && dist[t] == INF) return INF;
if (i) res += dist[t];

for (int j = 1; j <= n; j ++ ) dist[j] = min(dist[j], g[t][j]);

st[t] = true;
}

return res;
}

int main()
{
cin >> n >> m;

memset(g, 0x3f, sizeof g);

while (m -- )
{
int a, b, c;
cin >> a >> b >> c;
g[a][b] = g[b][a] = min(g[a][b], c);
}

int t = prim();

if (t == INF) puts("impossible");
else cout << t;

return 0;
}


yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 210, INF = 1e9;

int n, m, Q;
int d[N][N];

void floyd()
{
for (int k = 1; k <= n; k ++ )
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= n; j ++ )
d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

int main()
{
cin >> n >> m >> Q;

for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= n; j ++ )
if (i == j) d[i][j] = 0;
else d[i][j] = INF;

while (m -- )
{
int a, b, w;
cin >> a >> b >> w;

d[a][b] = min(d[a][b], w);
}

floyd();

while (Q -- )
{
int a, b;
cin >> a >> b;

if (d[a][b] > INF / 2) puts("impossible");
else cout << d[a][b] << endl;
}

return 0;
}


yolandaw
1个月前
模板级补全收录了floyd算法吗


yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 150010;

int n, m;
int h[N], e[N], w[N], ne[N], idx;
int dist[N], cnt[N];
bool st[N];

void add(int a, int b, int c)  // 添加一条边a->b，边权为c
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

int spfa()  // 求1号点到n号点的最短路距离，如果从1号点无法走到n号点则返回-1
{
queue<int> q;
for (int i = 1; i <= n; i ++ )
{
st[i] = true;
q.push(i);
}

while (q.size())
{
int t = q.front();
q.pop();
st[t] = false;

for (int i = h[t]; i != -1; i = ne[i])
{
int j = e[i];
if (dist[j] > dist[t] + w[i])
{
dist[j] = dist[t] + w[i];
cnt[j] = cnt[t] + 1;

if (cnt[j] >= n) return true;
if (!st[j])     // 如果队列中已存在j，则不需要将j重复插入
{
q.push(j);
st[j] = true;
}
}
}
}

return false;
}

int main()
{
cin >> n >> m;

memset(h, -1, sizeof h);

while (m -- )
{
int a, b, c;
cin >> a >> b >> c;
}

if (spfa()) puts("Yes");
else puts("No");

return 0;
}


yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 150010;

int n, m;
int h[N], e[N], w[N], ne[N], idx;
int dist[N];
bool st[N];

void add(int a, int b, int c)  // 添加一条边a->b，边权为c
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

int spfa()  // 求1号点到n号点的最短路距离，如果从1号点无法走到n号点则返回-1
{
queue<int> q;
memset(dist, 0x3f, sizeof dist);
dist[1] = 0;
q.push(1);
st[1] = true;

while (q.size())
{
int t = q.front();
q.pop();
st[t] = false;

for (int i = h[t]; i != -1; i = ne[i])
{
int j = e[i];
if (dist[j] > dist[t] + w[i])
{
dist[j] = dist[t] + w[i];
if (!st[j])     // 如果队列中已存在j，则不需要将j重复插入
{
q.push(j);
st[j] = true;
}
}
}
}

if (dist[n] == 0x3f3f3f3f) return -1;
return dist[n];
}

int main()
{
cin >> n >> m;

memset(h, -1, sizeof h);

while (m -- )
{
int a, b, c;
cin >> a >> b >> c;
}

int t = spfa();

if (t == -1) cout << "impossible";
else cout << t;

return 0;
}


yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 510, M = 10010;

int n, m, k;
int dist[N], bakcup[N];

struct Edge
{
int a, b, w;
}edges[M];

int bellman_ford()
{
memset(dist, 0x3f, sizeof dist);
dist[1] = 0;

for (int i = 0; i < k; i ++ )
{
memcpy(bakcup, dist, sizeof dist);
for (int j = 0; j < m; j ++ )
{
int a = edges[j].a, b = edges[j].b, w = edges[j].w;
dist[b] = min(dist[b], bakcup[a] + w);
}
}

if (dist[n] > 0x3f3f3f3f / 2) return -1;
return dist[n];
}

int main()
{
cin >> n >> m >> k;

for (int i = 0; i < m; i ++ )
{
int a, b, w;
cin >> a >> b >> w;
edges[i] = {a, b, w};
}

int t = bellman_ford();

if (t == -1) puts("impossible");
else cout << t;

return 0;
}


yolandaw
1个月前
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 150010;

int n, m;
int h[N], e[N], w[N], ne[N], idx;
int dist[N];
bool st[N];

void add(int a, int b, int c)  // 添加一条边a->b，边权为c
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

int dijkstra()  // 求1号点到n号点的最短路距离，如果从1号点无法走到n号点则返回-1
{
memset(dist, 0x3f, sizeof dist);
dist[1] = 0;
priority_queue<PII, vector<PII>, greater<PII>> heap;
heap.push({0, 1});

while (heap.size())
{
auto t = heap.top();
heap.pop();

int ver = t.second, distance = t.first;

if (st[ver]) continue;
st[ver] = true;

for (int i = h[ver]; i != -1; i = ne[i])
{
int j = e[i];
if (dist[j] > distance + w[i])
{
dist[j] = distance + w[i];
heap.push({dist[j], j});
}
}
}

if (dist[n] == 0x3f3f3f3f) return -1;
return dist[n];
}

int main()
{
cin >> n >> m;

memset(h, -1, sizeof h);

while (m -- )
{
int a, b, c;
cin >> a >> b >> c;