最短路径（DP的应用）
单源最短路径，不允许出现负环
核心思想：更新估算距离，松弛
$$\delta (u,v)\le \delta (u,x)+\delta (x,v)$$

时间复杂度与采用的数据结构有关，标准的dijkstra应该是用堆实现的。
Array O($v^2$)
Binary heap O($(V+E)lgV$)
Fibonacci heap O($E+VlgV$)

如果对于所有的边权值均为1，那么Dijkstra算法可以用BFS实现
使用FIFO队列代替Priority队列，其时间复杂度为O($V+E$)

数组实现：

#include <iostream>
using namespace std;
void dijkstra();
int e[10][10];
int vis[10];
int dis[10];
int n, m;
int min1 = 99999999;
int u = 0;
int main()
{cin >> n >> m;// 初始化邻接表for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){if (i == j){e[i][j] = 0;}else{e[i][j] = 99999999;}}}// 填充数据for (int i = 1; i <= m; i++){int a, b, c;cin >> a >> b >> c;e[a][b] = c;}for (int i = 1; i <= n; i++){dis[i] = e[1][i];}vis[1] = 1;dijkstra();for (int i = 1; i <= n; i++){cout << dis[i];}system("pause");return 0;
}
void dijkstra()
{for (int i = 1; i <= n - 1; i++){min1 = 99999999;// 寻找权值最小的点ufor (int j = 1; j <= n; j++){if (vis[j] == 0 && dis[j] < min1){min1 = dis[j];u = j;}}vis[u] = 1;for (int v = 1; v <= n; v++){// 对于每个u可达的v来说if (e[u][v] < 99999999){// 如果当前的dis[v]不满足三角形不等式，那么进行松弛操作if (dis[v] > dis[u] + e[u][v]){dis[v] = dis[u] + e[u][v];}}}}
}

堆实现

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int Ni = 10000;
const int INF = 1 << 27;
struct node
{int point, value;// 构造node(int a, int b){point = a;value = b;}// 重载<操作符bool operator<(const node &a) const{// 对小于运算符进行重载，如果两个值相等，那么继续判断point，如果不等，则返回falseif (value == a.value){return point < a.point;}else{return value > a.value;}}
};
vector<node> e[Ni];
int dis[Ni], n;
priority_queue<node> q;
void dijkstra();
int main()
{int a, b, c, m;scanf("%d%d", &n, &m);while (m--){scanf("%d%d%d", &a, &b, &c);e[a].push_back(node(b, c));e[b].push_back(node(a, c));}// for (int i = 0; i <= n; i++)// {//     dis[i] = INF;// }memset(dis, 0x3f, sizeof(dis));dis[1] = 0;// 优先队列，队头元素最大，但是如果类型为struct需要重载<运算符q.push(node(1, dis[1]));dijkstra();for (int i = 1; i <= n; i++){printf("%d ", dis[i]);}system("pause");return 0;
}
void dijkstra()
{while (!q.empty()){node x = q.top();q.pop();for (int i = 0; i < e[x.point].size(); i++){node y = e[x.point][i];if (dis[y.point] > dis[x.point] + y.value){dis[y.point] = dis[x.point] + y.value;q.push(node(y.point, dis[y.point]));}}}
}