最短路徑（DP的應(yīng)用）
單源最短路徑，不允許出現(xiàn)負(fù)環(huán)
核心思想：更新估算距離，松弛
$$\delta (u,v)\le \delta (u,x)+\delta (x,v)$$

時(shí)間復(fù)雜度與采用的數(shù)據(jù)結(jié)構(gòu)有關(guān)，標(biāo)準(zhǔn)的dijkstra應(yīng)該是用堆實(shí)現(xiàn)的。
Array O($v^2$)
Binary heap O($(V+E)lgV$)
Fibonacci heap O($E+VlgV$)

如果對(duì)于所有的邊權(quán)值均為1，那么Dijkstra算法可以用BFS實(shí)現(xiàn)
使用FIFO隊(duì)列代替Priority隊(duì)列，其時(shí)間復(fù)雜度為O($V+E$)

數(shù)組實(shí)現(xiàn)：

#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;// 初始化鄰接表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;}}}// 填充數(shù)據(jù)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;// 尋找權(quán)值最小的點(diǎn)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++){// 對(duì)于每個(gè)u可達(dá)的v來(lái)說(shuō)if (e[u][v] < 99999999){// 如果當(dāng)前的dis[v]不滿足三角形不等式，那么進(jìn)行松弛操作if (dis[v] > dis[u] + e[u][v]){dis[v] = dis[u] + e[u][v];}}}} }

堆實(shí)現(xiàn)

#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;// 構(gòu)造node(int a, int b){point = a;value = b;}// 重載<操作符bool operator<(const node &a) const{// 對(duì)小于運(yùn)算符進(jìn)行重載，如果兩個(gè)值相等，那么繼續(xù)判斷point，如果不等，則返回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;// 優(yōu)先隊(duì)列，隊(duì)頭元素最大，但是如果類(lèi)型為struct需要重載<運(yùn)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]));}}} }

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