poj 1129(ID-DFS四色定理)

Channel Allocation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11099		Accepted: 5700

Description

When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receiver has a strong signal. However, the channels used by each repeater must be carefully chosen so that nearby repeaters do not interfere with one another. This condition is satisfied if adjacent repeaters use different channels.

Since the radio frequency spectrum is a precious resource, the number of channels required by a given network of repeaters should be minimised. You have to write a program that reads in a description of a repeater network and determines the minimum number of channels required.

Input

The input consists of a number of maps of repeater networks. Each map begins with a line containing the number of repeaters. This is between 1 and 26, and the repeaters are referred to by consecutive upper-case letters of the alphabet starting with A. For example, ten repeaters would have the names A,B,C,...,I and J. A network with zero repeaters indicates the end of input.

Following the number of repeaters is a list of adjacency relationships. Each line has the form:

A:BCDH

which indicates that the repeaters B, C, D and H are adjacent to the repeater A. The first line describes those adjacent to repeater A, the second those adjacent to B, and so on for all of the repeaters. If a repeater is not adjacent to any other, its line has the form

A:

The repeaters are listed in alphabetical order.

Note that the adjacency is a symmetric relationship; if A is adjacent to B, then B is necessarily adjacent to A. Also, since the repeaters lie in a plane, the graph formed by connecting adjacent repeaters does not have any line segments that cross.

Output

For each map (except the final one with no repeaters), print a line containing the minumum number of channels needed so that no adjacent channels interfere. The sample output shows the format of this line. Take care that channels is in the singular form when only one channel is required.

Sample Input

2
A:
B:
4
A:BC
B:ACD
C:ABD
D:BC
4
A:BCD
B:ACD
C:ABD
D:ABC
0

Sample Output

1 channel needed.
3 channels needed.
4 channels needed. 

Source

Southern African 2001

ID-DFS求解這道題比較容易（interative deepening depth-first search)

第一次樣例測試失敗，是因爲沒有處理好輸入（本來應該是從line讀取字符，結果寫成了cin獲取字符）。

注意四色定理的直觀描述：

任意一個無飛地的地圖都可以用四種顏色染色，使得沒有兩個相鄰國家染的顏色相同。事實上實際的地圖往往是有飛地的，比如很多國家會有兩塊地方。

提交記錄：

1、Accepted！

/*Source Code

Problem: 1129		User: 775700879
Memory: 744K		Time: 0MS
Language: G++		Result: Accepted
Source Code*/
#include 
#include 
#include 
#include 
using namespace std;
int n;
int data[30][30];
char line[50];
int color[50];
bool dfs(int step, int used, int up) {
    if (used > up) return false;
    if (step == n) {
        return true;
    }
    int i, j;
    bool isused[50] = {0};
    for (i = 0; i < step; i++) {
        if (data[step][i] == true && color[i] != -1) isused[color[i]] = true;
    }
    for (i = 1; i <= used; i++) {
        if (!isused[i]) {
            color[step] = i;
            if(dfs(step+1, used, up)) return true;
            color[step] = 0;
        }
    }
    color[step] = used+1;
    if (dfs(step+1, used+1, up)) return true;
    color[step] = 0;
    return false;
}
int main() {
    int i, j;
    while (cin >> n) {
        if (n == 0) break;
        char ch;
        memset(data, 0, sizeof(data));
        for (i = 0; i < n; i++) {
            cin >> line;
            color[i] = -1;
            for (j = 2; line[j] != 0; j++) {
                //cin >> ch;
                data[i][line[j]-'A'] = 1;
            }
        }
        for (i = 1; i <= n; i++) {
            if (dfs(0, 0, i)) {
                cout << i << " channel";
                if (i != 1) cout << "s";
                cout << " needed." << endl;
                break;
            }
        }
    }
}