POJ 1129 Channel Allocation 图的染色 最大团
2016-02-11 15:09
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%%% http://www.cnblogs.com/zhj5chengfeng/archive/2013/07/29/3224092.html
题意:对图染色,要求相邻节点不能同色,问最多要多少种颜色。
其实应该是个平面图来着。。四色定理还没用呢。。不管那么多了。。
首先,最少需要颜色种数的瓶颈在什么地方?自然是完全图,两两相邻,那么完全图的颜色种数自然是完全图的点数。
那么对于一个普通的图呢?自然是其完全子图,最小要多少种颜色就是其最大团的点数。
Channel Allocation
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
Sample Output
题意:对图染色,要求相邻节点不能同色,问最多要多少种颜色。
其实应该是个平面图来着。。四色定理还没用呢。。不管那么多了。。
首先,最少需要颜色种数的瓶颈在什么地方?自然是完全图,两两相邻,那么完全图的颜色种数自然是完全图的点数。
那么对于一个普通的图呢?自然是其完全子图,最小要多少种颜色就是其最大团的点数。
#include <cstdio> #include <cstring> using namespace std; const int N = 105; int ans, f , set , a ; bool dfs(int sz, int dep) { if (!sz) if (dep > ans) return ans = dep, 1; else return 0; for (int i = 1; i <= sz; i++) { if (dep + sz - i + 1 <= ans) return 0; int u = set[dep][i]; if (dep + f[u] <= ans) return 0; int num = 0; for (int j = i + 1; j <= sz; j++) if (a[u][set[dep][j]]) set[dep + 1][++num] = set[dep][j]; if (dfs(num, dep + 1)) return 1; } return 0; } int main() { char str[32]; int n, k, i, j, sz, t, x, y; while (scanf("%d", &n) == 1 && n) { memset(a, 0, sizeof a); memset(f, 0, sizeof f); for (i = 1; i <= n; i++) { scanf("%s", str); for (j = 2; str[j]; j++) x = i, y = str[j] - 'A' + 1, a[x][y] = a[y][x] = 1; } ans = 0; for (i = n; i; i--) { sz = 0; for (j = i + 1; j <= n; j++) if (a[i][j]) set[1][++sz] = j; dfs(sz, 1); f[i] = ans; } if (ans == 1) puts("1 channel needed."); else printf("%d channels needed.\n", ans); } return 0; }
Channel Allocation
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 13675 | Accepted: 6984 |
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.
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