Dijkstra算法
算法流程:
(a) 初始化:用起點v到該頂點w的直接邊(弧)初始化最短路徑,否則設為∞;
(b) 從未求得最短路徑的終點中選擇路徑長度最小的終點u:即求得v到u的最短路徑;
(c) 修改最短路徑:計算u的鄰接點的最短路徑,若(v,…,u)+(u,w)<(v,…,w),則以(v,…,u,w)代替。
(d) 重復(b)-(c),直到求得v到其余所有頂點的最短路徑。
特點:總是按照從小到大的順序求得最短路徑。
假設一共有N個節點,出發結點為s,需要一個一維數組prev[N]來記錄前一個節點序號,一個一維數組dist[N]來記錄從原點到當前節點最短路徑(初始值為s到Vi的邊的權值,沒有則為+∞),一個二維數組weights[N][N]來記錄各點之間邊的權重,按以上流程更新prev[N]和dist[N]。
參考代碼:
#include <iostream> #include <cstdlib> using namespace std; void Dijkstra(int n,int s,int *dist,int *prev,int w[][4]) { int maxint = 65535; bool *visit = new bool[n]; for (int i = 0; i < n; i++) { dist[i] = w[s][i]; visit[i] = false; if (dist[i] != maxint) { prev[i] = s; } } dist[s] = 0; visit[s] = true; for (int i = 0; i < n; i++) { int temp = maxint; int u = s; for (int j = 0; j < n; j++) { if ((!visit[j]) && (dist[j] < temp)) { u = j; temp = dist[j]; } } visit[u] = true; for (int j = 0; j < n; j++) { if (!visit[j]) { int newdist = dist[u] + w[u][j]; if (newdist < dist[j]) { dist[j] = newdist; prev[j] = u; } } } } delete []visit; } int main() { int n,v,u; int weight[4][4]={ 0,2,65535,4, 2,0,3,65535, 65535,3,0,2, 4,65535,2,0 }; int q = 0; int way[4]; int dist[4]; int prev[4]; int s = 1; int d = 3; Dijkstra(4, s, dist, prev, weight); cout<<"The least distance from "<<s<<" to "<<d<<" is "<<dist[d]<<endl; int w = d; while (w != s) { way[q++] = prev[w]; w = prev[w]; } cout<<"The path is "; for (int j = q-1; j >= 0; j--) { cout<<way[j]<<" ->"; } cout<<d<<endl; return 0; }
Bellman-Ford算法
Bellman-Ford算法能在更普遍的情況下(存在負權邊)解決單源點最短路徑問題。對於給定的帶權(有向或無向)圖 G=(V,E),其源點為s,加權函數 w 是邊集 E 的映射。對圖G運行Bellman-Ford算法的結果是一個布爾值,表明圖中是否存在着一個從源點s可達的負權回路。若不存在這樣的回路,算法將給出從源點s到圖G的任意頂點v的最短路徑d[v]。
Bellman-Ford算法流程分為三個階段:
(1)初始化:將除源點外的所有頂點的最短距離估計值 d[v] ←+∞, d[s] ←0;
(2)迭代求解:反復對邊集E中的每條邊進行松弛操作,使得頂點集V中的每個頂點v的最短距離估計值逐步逼近其最短距離;(運行|v|-1次)
(3)檢驗負權回路:判斷邊集E中的每一條邊的兩個端點是否收斂。如果存在未收斂的頂點,則算法返回false,表明問題無解;否則算法返回true,並且從源點可達的頂點v的最短距離保存在 d[v]中。
算法描述如下:
Bellman-Ford(G,w,s) :boolean //圖G ,邊集 函數 w ,s為源點
1 for each vertex v ∈ V(G) do //初始化 1階段
2 d[v] ←+∞
3 d[s] ←0; //1階段結束
4 for i=1 to |v|-1 do //2階段開始,雙重循環。
5 for each edge(u,v) ∈E(G) do //邊集數組要用到,窮舉每條邊。
6 If d[v]> d[u]+ w(u,v) then //松弛判斷
7 d[v]=d[u]+w(u,v) //松弛操作 2階段結束
8 for each edge(u,v) ∈E(G) do
9 If d[v]> d[u]+ w(u,v) then
10 Exit false
11 Exit true
適用條件和范圍:
1.單源最短路徑(從源點s到其它所有頂點v);
2.有向圖&無向圖(無向圖可以看作(u,v),(v,u)同屬於邊集E的有向圖);
3.邊權可正可負(如有負權回路輸出錯誤提示);
4.差分約束系統;
#include <stdio.h> #include <stdlib.h> /* Let INFINITY be an integer value not likely to be confused with a real weight, even a negative one. */ #define INFINITY ((1 << 14)-1) typedef struct { int source; int dest; int weight; } Edge; void BellmanFord(Edge edges[], int edgecount, int nodecount, int source) { int *distance =(int*) malloc(nodecount*sizeof(int)); int i, j; for (i=0; i < nodecount; ++i) distance[i] = INFINITY; distance[source] = 0; for (i=0; i < nodecount; ++i) { int nbChanges = 0; for (j=0; j < edgecount; ++j) { if (distance[edges[j].source] != INFINITY) { int new_distance = distance[edges[j].source] + edges[j].weight; if (new_distance < distance[edges[j].dest]) { distance[edges[j].dest] = new_distance; nbChanges++; } } } // if one iteration had no impact, further iterations will have no impact either if (nbChanges == 0) break; } for (i=0; i < edgecount; ++i) { if (distance[edges[i].dest] > distance[edges[i].source] + edges[i].weight) { puts("Negative edge weight cycles detected!"); free(distance); return; } } for (i=0; i < nodecount; ++i) { printf("The shortest distance between nodes %d and %d is %d\n", source, i, distance[i]); } free(distance); return; } int main(void) { /* This test case should produce the distances 2, 4, 7, -2, and 0. */ Edge edges[10] = {{0,1, 5}, {0,2, 8}, {0,3, -4}, {1,0, -2}, {2,1, -3}, {2,3, 9}, {3,1, 7}, {3,4, 2}, {4,0, 6}, {4,2, 7}}; BellmanFord(edges, 10, 5, 4); return 0; }