POJ 3020 Antenna Placement 【最小邊覆蓋】
阿新 • • 發佈:2018-11-22
傳送門:http://poj.org/problem?id=3020
Antenna PlacementTime Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11098 | Accepted: 5464 |
Description
The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most striking reason why they got the job, is their discovery of a new, highly noise resistant, antenna. It is called 4DAir, and comes in four types. Each type can only transmit and receive signals in a direction aligned with a (slightly skewed) latitudinal and longitudinal grid, because of the interacting electromagnetic field of the earth. The four types correspond to antennas operating in the directions north, west, south, and east, respectively. Below is an example picture of places of interest, depicted by twelve small rings, and nine 4DAir antennas depicted by ellipses covering them.Obviously, it is desirable to use as few antennas as possible, but still provide coverage for each place of interest. We model the problem as follows: Let A be a rectangular matrix describing the surface of Sweden, where an entry of A either is a point of interest, which must be covered by at least one antenna, or empty space. Antennas can only be positioned at an entry in A. When an antenna is placed at row r and column c, this entry is considered covered, but also one of the neighbouring entries (c+1,r),(c,r+1),(c-1,r), or (c,r-1), is covered depending on the type chosen for this particular antenna. What is the least number of antennas for which there exists a placement in A such that all points of interest are covered?
Input
Output
Sample Input
2 7 9 ooo**oooo **oo*ooo* o*oo**o** ooooooooo *******oo o*o*oo*oo *******oo 10 1 * * * o * * * * * *
Sample Output
17 5
Source
Svenskt Mästerskap i Programmering/Norgesmesterskapet 2001
題意概括:
給一個高為 H 寬為 W 的圖案, “ * ” 表示城市,每個城市可以覆蓋 它 上下左右相鄰的 其中一個城市,問最好需要多少城市才能把所有城市覆蓋。
解題思路:
按照城市編號,然後拆點建圖,相鄰兩個城市可以覆蓋的就連邊,跑一遍最小邊覆蓋。
AC code:
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <cmath> 5 #define INF 0x3f3f3f3f 6 using namespace std; 7 const int MAXH = 45; 8 const int MAXW = 15; 9 const int MAXN = 500; 10 struct Edge 11 { 12 int v, nxt; 13 }edge[MAXN*MAXN]; 14 int head[MAXN], cnt; 15 int linker[MAXN]; 16 bool used[MAXN]; 17 char str[MAXH][MAXW]; 18 int a[MAXH][MAXW]; 19 int sum; 20 int H, W; 21 22 void init() 23 { 24 memset(head, -1, sizeof(head)); 25 memset(linker, -1, sizeof(linker)); 26 memset(a, 0, sizeof(a)); 27 cnt = 0; 28 sum = 0; 29 } 30 31 void add(int from, int to) 32 { 33 edge[cnt].v = to; 34 edge[cnt].nxt = head[from]; 35 head[from] = cnt++; 36 } 37 38 bool Find(int x) 39 { 40 int v; 41 for(int i = head[x]; i != -1; i = edge[i].nxt){ 42 v = edge[i].v; 43 if(!used[v]){ 44 used[v] = true; 45 if(linker[v] == -1 || Find(linker[v])){ 46 linker[v] = x; 47 return true; 48 } 49 } 50 } 51 return false; 52 } 53 54 int main() 55 { 56 int T_case; 57 scanf("%d", &T_case); 58 while(T_case--){ 59 init(); 60 scanf("%d%d", &H, &W); 61 for(int i = 0; i < H; i++){ 62 scanf("%s", &str[i]); 63 for(int j = 0; j < W; j++) 64 if(str[i][j]=='*') a[i][j] = ++sum; 65 } 66 67 for(int i = 0; i < H; i++){ 68 for(int j = 0; j < W; j++){ 69 if(a[i][j]){ 70 if(i > 0 && a[i-1][j]) add(a[i][j], a[i-1][j]); 71 if(i < H-1 && a[i+1][j]) add(a[i][j], a[i+1][j]); 72 if(j > 0 && a[i][j-1]) add(a[i][j], a[i][j-1]); 73 if(j < W-1 && a[i][j+1]) add(a[i][j], a[i][j+1]); 74 } 75 } 76 } 77 78 int ans = 0; 79 for(int i = 1; i <= sum; i++){ 80 memset(used, 0, sizeof(used)); 81 if(Find(i)) ans++; 82 } 83 printf("%d\n", sum-ans/2); 84 } 85 return 0; 86 }View Code