只能说构图是痛苦的。。。。。完全没思路,思路在下面。。。。
无向二分图的最小路径覆盖 = 顶点数 – 最大二分匹配数/2
http://user.qzone.qq.com/289065406/blog/1299322779
code
#include <set> #include <map> #include <ctime> #include <queue> #include <cmath> #include <stack> #include <limits> #include <vector> #include <bitset> #include <string> #include <cstdio> #include <cstring> #include <fstream> #include <string.h> #include <iostream> #include <algorithm> #define Si set<int> #define LL long long #define pb push_back #define PS printf(" ") #define Vi vector<int> #define LN printf("\n") #define lson l,m,rt << 1 #define rson m+1,r,rt<<1|1 #define SD(a) scanf("%d",&a) #define PD(a) printf("%d",a) #define SET(a,b) memset(a,b,sizeof(a)) #define FF(i,a) for(int i(0);i<(a);i++) #define FD(i,a) for(int i(a);i>=(1);i--) #define FOR(i,a,b) for(int i(a);i<=(b);i++) #define FOD(i,a,b) for(int i(a);i>=(b);i--) #define readf freopen("input.txt","r",stdin) #define writef freopen("output.txt","w",stdout) const int maxn = 502; const int INF = 0x3fffffff; const int dx[]={0,1,0,-1}; const int dy[]={1,0,-1,0}; const double pi = acos(-1.0); const double eps= 1e-7; using namespace std; int N,M,V1,V2,cnt,ans;//V1,V2是二分图的两个顶点集合 int city[maxn][maxn]; bool vis[maxn];//记录一边的顶点是否搜索过 bool Graph[maxn][maxn]; int link[maxn];//记录连接这个y顶点的x点 bool dfs(int x){ FOR(i,1,V2){ if(!vis[i] && Graph[x][i]){ vis[i]=true; if(link[i]==0 || dfs(link[i])){ link[i]=x; return true; } } } return false; } void getGraph(){ FOR(i,1,N) FOR(j,1,M){ if(city[i][j]){ FF(k,4){ int nx=i+dx[k]; int ny=j+dy[k]; if(city[nx][ny]){ Graph[city[i][j]][city[nx][ny]]=true; } } } } } int main() { int cas; SD(cas); char ch; while(cas--){ cnt=ans=0; SET(city,0);SET(Graph,0);SET(link,0); SD(N);SD(M); FOR(i,1,N) FOR(j,1,M){ scanf(" %c",&ch); if(ch=='*'){ city[i][j]=++cnt;//代表这个点有城市 } } getGraph(); V1=V2=cnt; FOR(i,1,V1){ SET(vis,false); if(dfs(i)) ans++; } PD(cnt-ans/2);LN; } return 0; }