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

As more and more transactions between companies and people are being carried out electronically over the Internet, secure communications have become an important concern. The Internet Cryptographic Protocol Company (ICPC) specializes in secure business-to-business transactions carried out over a network. The system developed by ICPC is peculiar in the way it is deployed in the network.

A network like the Internet can be modeled as a directed graph: nodes represent machines or routers, and edges correspond to direct connections, where data can be transmitted along the direction of an edge. For two nodes to communicate, they have to transmit their data along directed paths from the first node to the second, and from the second node to the first.

To perform a secure transaction, ICPC's system requires the installation of their software not only on the two endnodes that want to communicate, but also on all intermediate nodes on the two paths connecting the end-nodes. Since ICPC charges customers according to how many copies of their software have to be installed, it would be interesting to have a program that for any network and end-node pair finds the cheapest way to connect the nodes.

Input 

The input consists of several descriptions of networks. The first line of each description contains two integers N and M (2N100), the number of nodes and edges in the network, respectively. The nodes in the network are labeled 1, 2,..., N, where nodes 1 and 2 are the ones that want to communicate. The first line of the description is followed by M lines containing two integers X and Y (1X, YN), denoting that there is a directed edge from X to Y in the network.

The last description is followed by a line containing two zeroes.

Output 

For each network description in the input, display its number in the sequence of descriptions. Then display the minimum number of nodes on which the software has to be installed, such that there is a directed path from node 1 to node 2 using only the nodes with the software, and also a path from node 2 to node 1 with the same property. (Note that a node can be on both paths but a path need not contain all the nodes.) The count should include nodes 1 and 2.

If node 1 and 2 cannot communicate, display `Impossible' instead.

Follow the format in the sample given below, and display a blank line after each test case.

Sample Input 

8 12
1 3 
3 4 
4 2 
2 5 
5 6 
6 1 
1 7 
7 1 
8 7 
7 8 
8 2 
2 8 

2 1 
1 2 

0 0

Sample Output 

Network 1 
Minimum number of nodes = 4 

Network 2 
Impossible

Claimer: The data used in this problem is unofficial data prepared by Shahriar Manzoor. So any mistake here does not imply mistake in the offcial judge data. Only Shahriar Manzoor is responsible for the mistakes. Report mistakes to Shahriar_manzoor@yahoo.com

在有向圖中，假設節點是主機，那麼圖中經過的部分都要安裝軟體才能進行傳輸工作，問從點 1 傳到點 2，再傳回點 1，問其中最少安裝軟體的主機，點可以重覆走。
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解法挺妙的，設定 dp[a][b] 表示為 1 ->a, b -> 1 的最少兩條路徑，然後依序增點嘗試，最後答案為 dp[2][2]，但是路徑會回繞，因此特別考慮一種特殊型，使得 dp[b][a] = dp[a][b] + 造成 loop 的最小成本。

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
int g[105][105];
int dp[105][105], inq[105][105];
int main() {
    //freopen("in.txt","r+t",stdin);
    //freopen("out2.txt","w+t",stdout);
    int n, m;
    int i, j, k;
    int x, y, cost;
    int cases = 0;
    while(scanf("%d %d", &n, &m) == 2 && n) {
        memset(inq, 0, sizeof(inq));
        for(i = 1; i <= n; i++) {
            for(j = 1; j <= n; j++) {
                g[i][j] = 0xfffffff;
            }
        }
        for(i = 0; i < m; i++) {
            scanf("%d %d", &x, &y);
            g[x][y] = 1;
        }
        for(k = 1; k <= n; k++)
            for(i = 1; i <= n; i++)
                for(j = 1; j <= n; j++)
                    g[i][j] = min(g[i][j], g[i][k]+g[k][j]);

        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
                dp[i][j] = 0xfffffff;
        queue<int> X, Y;
        X.push(1), Y.push(1);
        dp[1][1] = 1;
        while(!X.empty()) {
            x = X.front(), X.pop();
            y = Y.front(), Y.pop();
            inq[x][y] = 0;
            //printf("%2d %2d %d\n", x, y, dp[x][y]);
            for(i = 1; i <= n; i++) {
                cost = (x != i && y != i);
                if(g[x][i] == 1) {
                    if(dp[i][y] > dp[x][y]+cost) {
                        dp[i][y] = dp[x][y]+cost;
                        if(inq[i][y] == 0) {
                            inq[i][y] = 1;
                            X.push(i), Y.push(y);
                        }
                    }
                }
                if(g[i][y] == 1) {
                    if(dp[x][i] > dp[x][y]+cost) {
                        dp[x][i] = dp[x][y]+cost;
                        if(inq[x][i] == 0) {
                            inq[x][i] = 1;
                            X.push(x), Y.push(i);
                        }
                    }
                }
            }
            if(x != y) {
                if(dp[y][x] > dp[x][y] + g[x][y]-1) {
                    dp[y][x] = dp[x][y] + g[x][y]-1;
                    if(inq[y][x] == 0) {
                        inq[y][x] = 1;
                        X.push(y), Y.push(x);
                    }
                }
            }
        }
        printf("Network %d\n", ++cases);
        if(dp[2][2] == 0xfffffff)
            puts("Impossible");
        else
            printf("Minimum number of nodes = %d\n", dp[2][2]);
        puts("");
    }
    return 0;
}
/*
11 46
2 4
3 11
10 4
11 11
3 10
9 9
9 7
10 1
6 8
4 9
11 1
6 2
3 9
1 11
5 3
9 6
8 1
3 6
6 1
4 6
4 10
4 4
3 3
7 10
7 6
1 9
7 9
1 8
9 10
4 5
5 2
10 10
1 3
8 9
1 4
5 4
1 7
2 11
5 5
5 11
11 8
10 2
8 4
6 3
9 3
7 1
*/