Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

In the 25th century, civilization is struck by a series of calamities that eventually led mankind to build walled cities interconnected by tunnel bridges to facilitate transportation. Each walled city possesses a unique ore required to build and repair all infrastructure including the tunnels. This material which when combined with other ores from all others cities form an almost indestructible material called ``oreon".

Outside the walled cities are uncivilized barbarians armed with antiquated but destructive weaponry which can effectively shoot down any air transport, but only damage and not penetrate tunnel bridges. Thus each city is interconnected to more than one city in order to have access redundancy in case one of its interconnecting tunnels is damaged.

If a tunnel is damaged, it becomes impassable and would require a substantial amount of ``oreon" to repair the damage. When a single city is made isolated, meaning all of its interconnections are damaged, ``oreon" cannot be manufactured which may lead to the eventual destruction of the wall fortifying the city. You, being the head of the homeland defense unit, are tasked to ensure that all cities remain accessible even by at least a single interconnecting tunnel at all times. Faced with only a limited manpower in the defense unit, you have to determine which tunnel to protect using the least number of people and ensure that no city will be isolated.

Figure 2 shows a map of the walled cities, their interconnecting tunnels and the number of security personnel.

Input 

The input will contain several test cases. The first line will indicate the number of test cases. Each test case begins with a number representing the number of walled cities. Cities are labeled alphabetically using the letters in the English alphabet. The subsequent lines contain the number of security personnel needed to protect the tunnel connecting each city to all other cities. A value of zero implies no security personnel needed since no tunnel exists. You are to output which tunnel should be protected and how many personnel are needed for each tunnel.

Output 

The output shows the tunnel connection which is named after the cities that it connects (in alphabetical order) and the number of personel needed to protect the tunnel. Order the records in increasing order of presonal. In case two tunnels have the same number of personel, write them in lexicographycal order.

Sample Input 

1 
6 
0, 8, 12, 0, 0, 7 
8, 0, 0, 3, 0, 0 
12, 0, 0, 0, 6, 0 
0, 3, 0, 0, 0, 4 
0, 0, 6, 0, 0, 5 
7, 0, 0, 4, 5, 0

Sample Output 

Case 1: 
B-D 3 
D-F 4 
E-F 5 
C-E 6 
A-F 7

---

做法 : 最小生成樹
題目比較麻煩點的就是輸出時要按照字典順序

#include <stdio.h>
#include <stdlib.h>
int P[30], R[30];
int MakeInit(int N) {
	int i;
	for(i = 0; i <= N; i++)
		P[i] = i, R[i] = 1;
}
int FindP(int x) {
	if(P[x] != x)
		P[x] = FindP(P[x]);
	return P[x];
} 
int Link(int x, int y) {
	if(R[x] > R[y])
		P[y] = x, R[x] += R[y];
	else
		P[x] = y, R[y] += R[x];
}
int Union(int x, int y) {
	x = FindP(x), y = FindP(y);
	if(x != y) {
		Link(x, y);
		return 1;
	}
	return 0;
}
typedef struct {
	int x, y, v;
}Path;
int cmp(const void *i, const void *j) {
	Path *a, *b;
	a = (Path *)i, b = (Path *)j;
	if(a->v != b->v)
		return a->v - b->v;
	if(a->x != b->x)
		return a->x - b->x;
	return a->y - b->y;
}
int main() {
	int T, N, C = 0, i, j, tmp;
	char cha;
	scanf("%d", &T);
	while(T--) {
		scanf("%d", &N);
		Path D[1000];
		int M = 0;
		for(i = 0; i < N; i++) {
			for(j = 0; j < N; j++) {
				scanf("%d%c", &tmp, &cha);
				if(i < j && tmp)
					D[M].x = i, D[M].y = j, D[M].v = tmp, M++;
			}
		}
		MakeInit(N);
		qsort(D, M, sizeof(Path), cmp);
		printf("Case %d:\n", ++C);
		for(i = 0; i < M; i++) {
			if(Union(D[i].x, D[i].y) == 1)
				printf("%c-%c %d\n", D[i].x+'A', D[i].y+'A', D[i].v);
		}
	}
    return 0;
}