POJ 3020——Antenna Placement——————【 最小路径覆盖、奇偶性建图】

Antenna Placement

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3020

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

On the first row of input is a single positive integer n, specifying the number of scenarios that follow. Each scenario begins with a row containing two positive integers h and w, with 1 <= h <= 40 and 0 < w <= 10. Thereafter is a matrix presented, describing the points of interest in Sweden in the form of h lines, each containing w characters from the set ['*','o']. A '*'-character symbolises a point of interest, whereas a 'o'-character represents open space.

Output

For each scenario, output the minimum number of antennas necessary to cover all '*'-entries in the scenario's matrix, on a row of its own.

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

题目大意：用1*2的长条覆盖图中的“*”。问最少需要多少个长条。

解题思路：最小路径覆盖：选择最少的边，让每个顶点都被选中，单独的结点可以作为一条路径。建图思路都是奇偶性建图。但是不同的是，这个需要把所有“*”都覆盖。那么我们考虑最小路径覆盖。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 660;
const int INF = 0x3f3f3f3f;
vector<int>G[maxn];
int Mx[maxn], My[maxn], dx[maxn], dy[maxn], used[maxn], dis;
char Map[maxn][maxn];
int lis[maxn][maxn];
bool SearchP(int _n){
    queue<int>Q;
    memset(dx,-1,sizeof(dx));
    memset(dy,-1,sizeof(dy));
    int dis = INF;
    for(int i = 1; i <= _n; i++){
        if(Mx[i] == -1){
            dx[i] = 0;
            Q.push(i);
        }
    }
    int v;
    while(!Q.empty()){
        int u = Q.front(); Q.pop();
        if(dx[u] > dis) break;
        for(int i = 0; i < G[u].size(); i++){
            v = G[u][i];
            if(dy[v] == -1){
                dy[v] = dx[u] + 1;
                if(My[v] == -1){
                    dis = dy[v];
                }else{
                    dx[My[v]] = dy[v] + 1;
                    Q.push(My[v]);
                }
            }
        }
    }
    return dis != INF;
}
int dfs(int u){
    int v;
    for(int i = 0; i < G[u].size(); i++){
        v = G[u][i];
        if(!used[v] && dy[v] == dx[u] + 1){
            used[v] = 1;
            if(My[v] != -1 && dy[v] == dis){
                continue;
            }
            if(My[v] == -1 || dfs(My[v])){
                Mx[u] = v;
                My[v] = u;
                return true;
            }
        }
    }
    return false;
}
int MaxMatch(int ln,int rn){
    int ret = 0;
    memset(Mx,-1,sizeof(Mx));
    memset(My,-1,sizeof(My));
    while(SearchP(ln)){
        memset(used,0,sizeof(used));
        for(int i = 1; i <= ln; i++){
            if(Mx[i] == -1 && dfs(i)){
                ret++;
            }
        }
    }
    return ret;
}
int main(){
    int T, cas = 0, n, m, N, M;
    scanf("%d",&T);
    while(T--){
        n =  m = 0;
        scanf("%d%d",&N,&M);
        for(int i = 0; i <= 210; i++){
            G[i].clear();
        }
        for(int i = 0; i <= M+1; i++){
            Map[N+1][i] = 'o';
            Map[0][i] = 'o';
        }
        for(int i = 0; i <= N+1; i++){
            Map[i][0] = 'o';
            Map[i][M+1] = 'o';
        }
        for(int i = 1; i <= N; i++){
            getchar();
            for(int j = 1; j <= M; j++){
                scanf("%c",&Map[i][j]);
                if(Map[i][j] == '*'){
                    if((i+j)%2 == 0){
                        ++n;
                        lis[i][j] = n;
                        if(Map[i-1][j] == '*'){
                            G[n].push_back(lis[i-1][j]);
                        }
                        if(Map[i][j-1] == '*'){
                            G[n].push_back(lis[i][j-1]);
                        }
                    }else{
                        ++m;
                        lis[i][j] = m;
                        if(Map[i-1][j] == '*'){
                            G[lis[i-1][j]].push_back(m);
                        }
                        if(Map[i][j-1] == '*'){
                            G[lis[i][j-1]].push_back(m);
                        }
                    }
                }
            }
        }
        int res = MaxMatch(n,m);
        printf("%d\n",n+m-res);
    }
    return 0;
}
/*
55
4 4
o***
oo*o
oooo
oooo

*/