以行列為點建圖，每個點（x,y) 對應(yīng)一條邊連接x,y。二分圖的最小點覆蓋=最大匹配

//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) {return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) {return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #elsefreopen("data.in","r",stdin);// freopen("d:\\out1.txt","w",stdout); #endif } int getch() {int ch;while((ch=getchar())!=EOF) {if(ch!=' '&&ch!='\n')return ch;}return EOF; }const int maxn = 1100; vector<int> g[maxn]; int n, m; void Init() {for(int i=1; i<=n; i++)g[i].clear(); }void add(int u, int v) {g[u].push_back(v); }int match[maxn]; int vis[maxn];bool dfs(int u) {vis[u] = true;for(int i = 0; i < g[u].size(); i++){int v = g[u][i];int w = match[v];if(w<0||!vis[w]&&dfs(w)){match[u] = v;match[v] = u;return true;}}return false; } int solve() {memset(match,-1,sizeof(match));int res = 0;for(int u=1; u <= n; u++){if(match[u]<0){memset(vis, 0, sizeof(vis));if(dfs(u))res++;}}return res; } int main() {debug();while(scanf("%d%d", &n, &m) != EOF){for(int i=1; i<=m; i++){int u,v;scanf("%d%d", &u, &v);add(u, v+n);add(v+n, u);}printf("%d\n",solve());}return 0; } View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/BMan/p/3713655.html

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