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 computational complexity theory, polynomial-time reduction is an important concept.

If the existence of a polynomial-time algorithm for problem B implies that problem A also has a polynomial-time algorithm, we say that problem A has a polynomial-time reduction to problem B . The relation of reduction is transitive, i.e. if A has a reduction to B and B has a reduction to C , then A has a reduction to C .

Theoretical computer science researchers have found hundreds of polynomial-time reductions between problems, which build a large network of reductions in computational complexity theory.

In this network, some reductions are explicitly presented and others exist implicitly. For example, if the reductions from A to B and from B to C are explicitly presented and the reduction from A to C is not explicitly presented, we say that the reduction from A to C exists implicitly.

In fact, some reductions are not necessary to present explicitly in the network. For example, if reductions from A to B and from B to C exist explicitly or implicitly in the network, then the reduction from A to C is not necessary to present explicitly. By this fact, it's possible to simplify the network of reductions.

Your task is just to simplify the network of reductions such that the number of reductions explicitly presented in the simplified network is minimized and reduction from problem A to problem B exists explicitly or implicitly in the simplified network if and only if it exists in the original network explicitly or implicitly.

Input 

The input consists of multiple test cases. Each test case starts with a line containing two integers N and M (1N100, 0M10000) , which are the number of problems and the number of explicitly presented reductions in the network. The problems are numbered from 1 to N .

Each of the following M lines contains two integers A and B (1AN, 1BN, A B) , which means a polynomial-time reduction from problem A to problem B is explicitly presented in the network. The last test case is followed by a line containing two zeros.

Output 

For each test case, print a line containing the test case number (beginning with 1) followed by a integer which is the number of explicitly presented reductions in the simplified network.

Sample Input 

3 3 
1 2 
2 3 
1 3 
4 6 
1 2 
2 1 
2 3 
3 2 
3 4 
4 3 
0 0

Sample Output 

Case 1: 2 
Case 2: 4

---

看到 reduction 詞時，不經想到正規語言上到的證明 orz。
有 2 個問題 A, B，如果能在多項式時間內將 A 轉換成 B，
進而可以使用 B 的演算法解決，稱為 A reduce to B。
順道複習一下，忘得一乾二淨了！
題目描述為一個 reduce 關係，因此可以找到一些遞移關係，
ex. A -> B -> C 可以得到 A -> C。
化簡整張圖，使用最少(從原本存在扣除)的 explicitly 表示。
-----
用 SCC 縮圖，每個 SCC 元件中只保留環狀邊即可。
縮圖後做一次遞移閉包，找到必需的邊。
即 if i->j necessary, not exists i->k->j (k != i, k !=j)

#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <deque>
#include <algorithm>
using namespace std;
vector<int> g[105], vg[105], ng[105];
int vfind[105], findIdx;
int stk[105], stkIdx;
int in_stk[105], visited[105];
int contract[105];
//
int ret;
int scc(int nd) {
    in_stk[nd] = visited[nd] = 1;
    stk[++stkIdx] = nd;
    vfind[nd] = ++findIdx;
    int mn = vfind[nd], i;
    for(i = 0; i < g[nd].size(); i++) {
        if(!visited[g[nd][i]])
            mn = min(mn, scc(g[nd][i]));
        if(in_stk[g[nd][i]])
            mn = min(mn, vfind[g[nd][i]]);
    }
    if(mn == vfind[nd]) {
    	int cnt = 0;
        do {
            in_stk[stk[stkIdx]] = 0;
            contract[stk[stkIdx]] = nd;
            cnt++;
        } while(stk[stkIdx--] != nd);
        if(cnt != 1)	ret += cnt;
    }
    return mn;
} 
int main() {
	int cases = 0;
	int n, m;
	int i, j, k, x, y;
	while(scanf("%d %d", &n, &m) == 2 && n+m) {
		for(i = 1; i <= n; i++)
			g[i].clear();	
		for(i = 0; i < m; i++) {
			scanf("%d %d", &x, &y);
			g[x].push_back(y);
		}
		ret = 0;// answer.
        memset(visited, 0, sizeof(visited));
        memset(in_stk, 0, sizeof(in_stk));
        for(i = 1; i <= n; i++) {
        	if(visited[i] == 0) {
        		findIdx = stkIdx = 0;
        		scc(i);
        	}
        }
        int tr[105][105] = {};// transitive graph
        for(i = 1; i <= n; i++) {// new DAG
        	x = contract[i];
        	for(j = 0; j < g[i].size(); j++) {
        		y = contract[g[i][j]];
        		if(x != y)
        			tr[x][y] = 1;
        	}
        }
        // floyd
        for(k = 1; k <= n; k++) {
        	for(i = 1; i <= n; i++) {
        		for(j = 1; j <= n; j++)
        			tr[i][j] |= tr[i][k]&tr[k][j];
        	}
        }
        // count necessary edge
        for(i = 1; i <= n; i++) {
        	for(j = 1; j <= n; j++) {
        		if(i == j) 	continue;
        		if(tr[i][j])  {
        			int necessary = 1;
        			for(k = 1; k <= n; k++) {
        				if(k == i || k == j)
        					continue;
        				if(tr[i][k] && tr[k][j]) 
        					necessary = 0, k = n+1;
        			}
        			ret += necessary;
        		}
        	}
        }
        printf("Case %d: %d\n", ++cases, ret);
	}
	return 0;
}