136 lines
4.2 KiB
C
136 lines
4.2 KiB
C
/*
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Program: all-pairs-shortest-path-v1.c (Report comments/bugs to chikh@yuntech.edu.tw)
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Function: 實作Floyd-Warshall演算法，以書附的shortestPath.dat檔案作為輸入
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Notes:
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1) Floyd-Warshall演算法為「動態規劃」(dynamic programming)法
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2) 讀取資料所用程式碼取自書附的shortestPath.c但調整函式的原型(prototype)且避用全域變數
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3) 部分參考 https://www.geeksforgeeks.org/floyd-warshall-algorithm-dp-16/
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與 https://www.programiz.com/dsa/floyd-warshall-algorithm
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*/
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#include <stdio.h>
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#include <stdlib.h> /* for exit() */
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#define MAX_V 100 /* 最大頂點數 */
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#define Inf 1073741823 /* 此為2^30-1的大數字，為本程式假設的無限大值；可改用limits.h定義的INT_MAX */
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void init(int A[][MAX_V], int Y[][MAX_V], int *N) /* 讀取輸入資料檔、設定相鄰矩陣，把相關資料結構作初始化 */
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{
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FILE *fptr;
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int i, j;
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int weight;
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if ((fptr=fopen("shortestPath.dat","r")) == NULL) { /* 開啟資料檔 */
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perror("shortestPath.dat");
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exit(1);
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}
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fscanf(fptr,"%d",N); /* 讀取圖形節點數 */
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for (i = 1; i <= *N; i++ )
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for (j = 1; j <= *N; j++) {
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A[i][j] = (i==j)? 0: Inf; /* 初始化A[1..N][1..N]相鄰矩陣，所有元素均先預設為∞ */
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Y[i][j] = (i==j)? i: -1;
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}
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while (fscanf(fptr,"%d %d %d",&i,&j,&weight) != EOF) {
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A[i][j] = weight; /* 讀取連接頂點i與j的邊的權重值 */
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Y[i][j] = i; /* Y[i][j]預設為i */
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}
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fclose(fptr);
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}
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/* 底下floydWarshall()函式就每二頂點(i,j)之間計算最短距離值，並記錄可經由哪些中繼頂點達到最短 */
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void floydWarshall(int A[][MAX_V], int Y[][MAX_V], int N)
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{
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int i, j, k;
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for (k = 1; k <= N; k++)
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for (i = 1; i < N; i++)
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for (j = 1; j <= N; j++)
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if (A[i][k] + A[k][j] < A[i][j]) { //if (A[i][k] != Inf && A[k][j] != Inf && A[i][k] + A[k][j] < A[i][j]) {
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A[i][j] = A[i][k] + A[k][j];
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Y[i][j] = Y[k][j]; /* 不寫為 Y[i][j] = k */
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}
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/* 上面二層for迴圈用於修正每二頂點(i,j)之間的距離值，檢視有否經由k(任一個可能的中繼點)能得到更小距離值 */
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/*
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留意line 53的寫法，不寫為 Y[i][j] = k，同學可區分有何差別嗎?!
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理由：由line 37可知，若二頂點(i,j)之間有邊存在，將把Y[i][j]預設為i。
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因此，在line 53中，若k與j有邊存在，寫為Y[k][j]與寫為k的意義相同，意為到j之前的中繼點為k。
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但若k與j之間無直接的邊連接，則設Y[i][j] = Y[k][j]，意為i到j的最後一個中繼點將由從k到j的路徑所記錄的最後
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一個中繼點決定(Y[k][j]紀錄目前所知k到j距離j最近的中繼點)。
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透過line 53設置Y矩陣內容，將可回溯得到最短路徑
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*/
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}
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void printMatrix(int A[][MAX_V], int N)
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{
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int i, j;
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printf("\n\n運算後得頂點(i,j)間之矩陣資訊如下：\n ");
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for (i = 1; i <= N; i++) printf("%4d",i);
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for (i = 0; i <= N; i++) printf("%4s",i==0? "\n +":"----");
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for (i = 1; i <= N; i++) {
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printf("\n頂點%d |",i);
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for (j = 1; j <= N; j++) {
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if (A[i][j] == Inf)
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printf("%5s","∞"); /* somewhat weird to use %5s instead of %4s here */
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else
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printf("%4d",A[i][j]);
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}
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//printf("\n");
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}
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}
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int backtrack(int A[][MAX_V], int Y[][MAX_V], int source, int dest)
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{
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int distance, k = Y[source][dest]; /* k為source到dest之間的中繼頂點 */
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if (source == dest) {
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printf("頂點%d →",source);
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return A[source][dest];
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}
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if (Y[source][dest] == -1) {
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printf("頂點%d與%d之間沒有路徑存在:( ",source,dest);
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return Inf;
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}
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distance = backtrack(A,Y,source,k);
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printf(" %d →",dest);
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return distance+A[k][dest];
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}
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void traverseMatrix(int A[][MAX_V],int Y[][MAX_V])
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{
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int distance, source, sink;
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printf("\n\n查詢任二點之間的最短路徑\n源點(0退出查詢) => ");
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scanf("%d",&source);
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if (source == 0) return;
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printf("終點(0退出查詢) => ");
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scanf("%d",&sink);
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if (sink == 0) return;
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printf("最短路徑：");
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distance = backtrack(A,Y,source,sink);
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printf("\b\b (所經之邊距離總和=%d)(查表A[%d][%d]=%d)",distance,source,sink,A[source][sink]); /* 抹除最後顯示多餘的"→" */
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}
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int main()
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{
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int N, A[MAX_V][MAX_V], Y[MAX_V][MAX_V] = {0};
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/* N為圖所含的頂點實際總數；A[1..N][1..N]為圖形的相鄰矩陣 */
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/* Y[1..N][1..N]記錄任二頂點(i,j)之間距j最近的中繼點，將用於最短路徑回溯之用 */
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printf("\n*** 解 all-pairs shortest-path 問題 ***");
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init(A,Y,&N); /* 讀取輸入資料檔、設定相鄰矩陣，並把相關資料結構作初始化 */
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floydWarshall(A,Y,N);
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printMatrix(A,N);
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//printMatrix(Y,N);
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traverseMatrix(A,Y);
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return 0;
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}
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