解题报告

题目传送门

题意：

在h×w的矩阵中，o表示空地，*表示城市，无线设备只能装在城市上，要使城市全都覆盖需要多少设备。每个设备只能覆盖相邻的两个设备。

思路：

感觉是直接的最大匹配，求出两两匹配的最大数，加上没有匹配的城市就是要的答案。

网上看了题解，正解是最小路径覆盖。

最小路径覆盖＝｜G｜－最大匹配数

在一个N*N的有向图中，路径覆盖就是在图中找一些路经，使之覆盖了图中的所有顶点，
且任何一个顶点有且只有一条路径与之关联；（如果把这些路径中的每条路径从它的起始点走到它的终点，
那么恰好可以经过图中的每个顶点一次且仅一次）；如果不考虑图中存在回路，那么每每条路径就是一个弱连通子集．

我的最大匹配过的。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
struct node {
    int v,w,next;
} edge[3000];
int cnt,k,head[500],vis[500],pre[500],mmap[100][100],h,w;
int dx[]= {-1,0,1,0};
int dy[]= {0,1,0,-1};
void add(int u,int v,int w) {
    edge[cnt].v=v;
    edge[cnt].w=w;
    edge[cnt].next=head[u];
    head[u]=cnt++;
    edge[cnt].v=u;
    edge[cnt].w=w;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}
int dfs(int x) {
    for(int i=head[x]; i!=-1; i=edge[i].next) {
        int v=edge[i].v;
        if(!vis[v]) {
            vis[v]=1;
            if(pre[v]==-1||dfs(pre[v])) {
                pre[v]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main() {
    int t,i,j;
    char str[100];
    //freopen("a.in","r",stdin);
    scanf("%d",&t);
    while(t--) {
        memset(head,-1,sizeof(head));
        memset(edge,0,sizeof(edge));
        memset(pre,-1,sizeof(pre));
        memset(mmap,0,sizeof(mmap));
        k=0,cnt=0;
        scanf("%d%d",&h,&w);
        for(i=1; i<=h; i++) {
            scanf("%s",str);
            for(j=0; j<w; j++) {
                if(str[j]=='*')
                    mmap[i][j+1]=++k;
            }
        }
        for(i=1; i<=h; i++) {
            for(j=1; j<=w; j++) {
                if(mmap[i][j]) {
                    for(int l=0; l<4; l++) {
                        if(mmap[i+dx[l]][j+dy[l]])
                            add(mmap[i][j],mmap[i+dx[l]][j+dy[l]],1);
                    }
                }
            }
        }
        int ans=0;
        for(i=1; i<=k; i++) {
            memset(vis,0,sizeof(vis));
            ans+=dfs(i);
        }
        printf("%d\n",k-(ans/2)*2+(ans/2));
    }
}

<pre name="code" class="cpp">官方测试数据：
input：
8
40 10
**********
**********
**********
**********
**********
**********
**********
**********
**********
**********
**********
**o*******
**oo******
**********
**********
**********
**********
**********
**********
**********
****oo****
**********
**********
**********
**********
**********
*****ooo**
*****o*o**
*****ooo**
**********
**********
**********
**********
**********
**********
**********
**********
**********
**********
**********
7 9
ooo**oooo
**oo*ooo*
o*oo**o**
ooooooooo
*******oo
o*o*oo*oo
*******oo
10 1
*
*
*
o
*
*
*
*
*
*
2 1
o
o
40 10
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*ooooooo*o
o*o***o*o*
*oo*o*oo*o
o*o***o*o*
*o*ooo*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
o*o*o*o*o*
*o*o*o*o*o
37 10
oooooooooo
o**ooooooo
o*oo**oooo
o*****oooo
oo**oooooo
oooooooooo
o**ooooooo
o*o**ooooo
o******ooo
oooooooooo
o***oooooo
o**ooooooo
oooooo***o
o*********
o***ooo**o
oo*oooo**o
oo****oo*o
oooooooooo
oooooooooo
oooooooooo
oooo****oo
ooo**oo**o
oooo****oo
**oooooooo
oooooooooo
oooo*ooooo
oo****oooo
oooo**oooo
oooo*ooo*o
o*****o**o
o*o******o
ooooo*oo*o
o**oo*oo**
oo****oo*o
ooo***oooo
ooo*****oo
oooooooooo
9 10
**********
*oooooooo*
*o******o*
*o*oooo*o*
*o*oooo*o*
*o*oooo*o*
*o******o*
*oooooooo*
**********
2 1
*
*

output：
 195
 17
 5
 0
 193
 58
 26
 1

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