Dijkstra
AcWing 849. Dijkstra求最短路 I
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 510;
int n, m;
int g[N][N];
int dist[N];
bool st[N];
int dijkstra()
{
memset(dist, 0x3f, sizeof dist);
dist[1] = 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;
st[t] = true;
for (int j = 1; j <= n; j++)
dist[j] = min(dist[j], dist[t] + g[t][j]);
}
if (dist[n] == 0x3f3f3f3f) return -1; //这个地方== 被打成了=,调试了半天
else dist[n];
}
int main()
{
scanf("%d%d", &n, &m);
// for (int i = 1; i <= n; i++)
// for (int j = 1; j <=n; j++)
// if (i == j) g[i][j] = 0;
// else g[i][j]= INF;
memset(g, 0x3f, sizeof g);
while (m--)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
g[a][b] = min(g[a][b], c); //重边,邻接矩阵中只存最小的权值
}
int t = dijkstra();
printf("%d\n", t);
return 0;
}
AcWing 850. Dijkstra求最短路 II
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 100010;
int n, m;
int h[N], w[N], e[N], ne[N], idx;
int dist[N];
bool st[N];
void add(int a, int b, int c)
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
int dijkstra()
{
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;
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; //==, = 再一次打错
return dist[n];
}
int main()
{
scanf("%d%d", &n, &m);
memset(h, -1, sizeof h);
while (m--)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c);
}
int t = dijkstra();
printf("%d\n", t);
return 0;
}