題目鏈接：點擊查看

題目大意：給出一張無向圖，要求將其分為兩個集合，使得最小割最小

題目分析：算法學習自：https://blog.csdn.net/dingdi3021/article/details/101960508

全局最小割，用類最大生成樹實現的，時間復雜度 $O(n^3)$

代碼：

// Problem: Minimum Cut // Contest: Virtual Judge - POJ // URL: https://vjudge.net/problem/POJ-2914 // Memory Limit: 65 MB // Time Limit: 10000 ms // // Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2) // #pragma GCC optimize("Ofast","inline","-ffast-math") // #pragma GCC target("avx,sse2,sse3,sse4,mmx") #include<iostream> #include<cstdio> #include<string> #include<ctime> #include<cmath> #include<cstring> #include<algorithm> #include<stack> #include<climits> #include<queue> #include<map> #include<set> #include<sstream> #include<cassert> #include<bitset> #include<list> #define lowbit(x) x&-x using namespace std; typedef long long LL; typedef unsigned long long ull; template<typename T> inline void read(T &x) {T f=1;x=0;char ch=getchar();while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();x*=f; } template<typename T> inline void write(T x) {if(x<0){x=~(x-1);putchar('-');}if(x>9)write(x/10);putchar(x%10+'0'); } const int inf=0x3f3f3f3f; const int N=510; int maze[N][N]; int v[N],wg[N],vis[N]; int SW(int n) {int res = -1;for(int i = 0; i < n; ++i) v[i] = i; //點標號的馬甲，一開始都是自己。while(n > 1){memset(vis,0,sizeof(vis)); //標記數組memset(wg,0,sizeof(wg)); //權和數組int pre = 0; //起點為0號點vis[pre] = 1; //標記起點，雖無實際用處，僅為提醒自己for(int i = 1; i < n; ++i){int p = -1;for(int j = 1; j < n; ++j) if(!vis[v[j]]){ //尋找下個拓展點wg[v[j]] += maze[v[pre]][v[j]]; //下個點的權和加上邊權if(p == -1 || wg[v[j]] > wg[v[p]]) p = j; //尋找wg值最大的點}vis[v[p]] = 1; //標記下個點已經遍歷if(i == n - 1){ //最后個點，需要合并if(res == -1) res = wg[v[p]];else res = min(res,wg[v[p]]); //更新res，取最小割for(int j = 0; j < n; ++j){maze[v[pre]][v[j]] += maze[v[p]][v[j]];maze[v[j]][v[pre]] += maze[v[j]][v[p]];}v[p] = v[--n];}pre = p;}}return res; } int main() { #ifndef ONLINE_JUDGE // freopen("data.in.txt","r",stdin); // freopen("data.out.txt","w",stdout); #endif // ios::sync_with_stdio(false);int n,m;while(scanf("%d%d",&n,&m)!=EOF) {memset(maze,0,sizeof(maze));while(m--) {int u,v,w;read(u),read(v),read(w);maze[u][v]+=w;maze[v][u]+=w;}printf("%d\n",SW(n));}return 0; }

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