吝嗇的國度
- 描述
- 在一個吝嗇的國度裏有N個城市,這N個城市間只有N-1條路把這個N個城市連接起來。現在,Tom在第S號城市,他有張該國地圖,他想知道如果自己要去參觀第T號城市,必須經過的前一個城市是幾號城市(假設你不走重複的路)。
- 輸入
- 第一行輸入一個整數M表示測試數據共有M(1<=M<=5)組
每組測試數據的第一行輸入一個正整數N(1<=N<=100000)和一個正整數S(1<=S<=100000),N表示城市的總個數,S表示參觀者所在城市的編號
隨後的N-1行,每行有兩個正整數a,b(1<=a,b<=N),表示第a號城市和第b號城市之間有一條路連通。 - 輸出
- 每組測試數據輸N個正整數,其中,第i個數表示從S走到i號城市,必須要經過的上一個城市的編號。(其中i=S時,請輸出-1)
- 樣例輸入
-
1 10 1 1 9 1 8 8 10 10 3 8 6 1 2 10 4 9 5 3 7
- 樣例輸出
-
-1 1 10 10 9 8 3 1 1 8
很明顯,這個些城市和道路構成了一個極小連通子圖,也就是生成樹。而出發的城市S就相當於這棵樹的樹根,一種較易想到的解法就是從出發城市開始,對整個地圖進行深度搜索,過程中記錄下前一個城市的編號,如下,採用鄰接表存儲地圖:
- #include <stdio.h>
- struct node
- {
- int num;
- node *next;
- };
- struct data_type
- {
- int priorCity;
- node *linkedCity;
- }map[100005];
- void MyDelete(int cityNum)
- {
- int i;
- node *p, *q;
- for (i = 1; i <= cityNum; i++)
- {
- p = map[i].linkedCity;
- while (p != NULL)
- {
- q = p->next;
- delete p;
- p = q;
- }
- }
- }
- void Travel(int currentCity, int priorCity)
- {
- map[currentCity].priorCity = priorCity;
- node *linkedCity = map[currentCity].linkedCity;
- while (linkedCity != NULL)
- {
- if (map[linkedCity->num].priorCity == 0)
- {
- Travel(linkedCity->num, currentCity);
- }
- linkedCity = linkedCity->next;
- }
- }
- int main()
- {
- int i, testNum, cityNum, startCity, cityA, cityB;
- node *p;
- scanf("%d", &testNum);
- while (testNum-- != 0)
- {
- scanf("%d%d", &cityNum, &startCity);
- for (i = 0; i <= cityNum; i++)
- {
- map[i].priorCity = 0;
- map[i].linkedCity = NULL;
- }
- for (i = 1; i < cityNum; i++)
- {
- scanf("%d%d", &cityA, &cityB);
- p = new node;
- p->num = cityB;
- p->next = map[cityA].linkedCity;
- map[cityA].linkedCity = p;
- p = new node;
- p->num = cityA;
- p->next = map[cityB].linkedCity;
- map[cityB].linkedCity = p;
- }
- Travel(startCity, -1);
- for (i = 1; i < cityNum; i++)
- {
- printf("%d ", map[i].priorCity);
- }
- printf("%d\n", map[i].priorCity);
- MyDelete(cityNum);
- }
- return 0;
- }
上面的地圖相當於一個無向圖,而在深度搜索時,需要的只是一個以出發城市爲中心,向四周輻射的有向圖。改進算法是在輸入數據的同時,就進行搜索地圖,因爲數據輸入未完成,所以輸入時得到的是一個子圖,這個子圖分兩種情況,一種是子圖中包含出發城市,子圖是一個有向圖,所以可以根據輸入的兩個城市哪一個離出發城市更近,確定結果;另一種子圖中不包含出發城市,此時,無法確定哪個城市離出發城市更近,所以先用鄰接表將這個無向子圖存儲起來,等到它與出發城市相連時,在對這個子圖進行深度搜索。
例如輸入的測試數據爲:10 18 1010 33 710 41 91 88 61 29 5則首先得到一個不包含出發城市的子圖:
在輸入數據1 8時之後,上面的子圖與出發城市相連,圖中紅色方塊代表出發城市,虛線箭頭代表並未在鄰接表中建立此聯繫:
- #include <stdio.h>
- struct node
- {
- int num;
- node *next;
- };
- struct data_type
- {
- int priorCity;
- bool start;
- node *linkedCity;
- }map[100005];
- void InitMap(int cityNum, int startCity)
- {
- int i;
- for (i = 0; i <= cityNum; i++)
- {
- map[i].priorCity = 0;
- map[i].start = false;
- map[i].linkedCity = NULL;
- }
- map[startCity].start = true;
- map[startCity].priorCity = -1;
- }
- void MyDelete(int cityNum)
- {
- int i;
- node *p, *q;
- for (i = 1; i <= cityNum; i++)
- {
- p = map[i].linkedCity;
- while (p != NULL)
- {
- q = p->next;
- delete p;
- p = q;
- }
- }
- }
- void Travel(int currentCity, int priorCity)
- {
- map[currentCity].priorCity = priorCity;
- map[currentCity].start = true;
- node *linkedCity = map[currentCity].linkedCity;
- while (linkedCity != NULL)
- {
- if (map[linkedCity->num].priorCity == 0)
- {
- Travel(linkedCity->num, currentCity);
- }
- linkedCity = linkedCity->next;
- }
- }
- int main()
- {
- int i, testNum, cityNum, startCity, cityA, cityB;
- node *p;
- scanf("%d", &testNum);
- while (testNum-- != 0)
- {
- scanf("%d%d", &cityNum, &startCity);
- InitMap(cityNum, startCity);
- for (i = 1; i < cityNum; i++)
- {
- scanf("%d%d", &cityA, &cityB);
- if (map[cityA].start)
- {
- if (map[cityB].linkedCity != NULL)
- {
- Travel(cityB, cityA);
- }
- else
- {
- map[cityB].priorCity = cityA;
- map[cityB].start = true;
- }
- }
- else if (map[cityB].start)
- {
- if (map[cityA].linkedCity != NULL)
- {
- Travel(cityA, cityB);
- }
- else
- {
- map[cityA].priorCity = cityB;
- map[cityA].start = true;
- }
- }
- else
- {
- p = new node;
- p->num = cityB;
- p->next = map[cityA].linkedCity;
- map[cityA].linkedCity = p;
- p = new node;
- p->num = cityA;
- p->next = map[cityB].linkedCity;
- map[cityB].linkedCity = p;
- }
- }
- for (i = 1; i < cityNum; i++)
- {
- printf("%d ", map[i].priorCity);
- }
- printf("%d\n", map[i].priorCity);
- MyDelete(cityNum);
- }
- return 0;
- }