标签:cond lin main iterator txt clu gre bitset typedef
题意: 一个有向图,存在从某个点为根的,可以到达所有点的一个最小生成树,则它就是最小树形图。
题目就是求这个最小的树形图。
参考资料:https://blog.csdn.net/txl199106/article/details/62045479
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <string> #include <vector> #include <map> #include <set> #include <queue> #include <list> #include <cstdlib> #include <iterator> #include <cmath> #include <iomanip> #include <bitset> #include <cctype> using namespace std; //#pragma comment(linker, "/STACK:102400000,102400000") //c++ #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second #define endl ‘\n‘ #define OKC ios::sync_with_stdio(false);cin.tie(0);cout.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar(); while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar(); return x=f?-x:x; } // #define _DEBUG; //*// #ifdef _DEBUG freopen("input", "r", stdin); // freopen("output.txt", "w", stdout); #endif /*-----------------show time----------------*/ int n,m; struct nn { int x,y; }a[110]; struct node { int u,v; double dis; }e[10009]; double get(int i,int j){ return sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x) + (a[i].y-a[j].y)*(a[i].y-a[j].y)); } double in[110]; int id[110],vis[110]; int pre[110]; int flag = 1; double zhuliu(int root){ double ans = 0.0; while(true){ for(int i=1; i<=n; i++){ in[i] = 9999999999.99; } for(int i=1; i<=m; i++){ node & b = e[i]; if(b.u!=b.v && in[b.v] > e[i].dis){ in[b.v] = e[i].dis; pre[b.v] = b.u; } } for(int i=1; i<=n; i++){ if(i!=root&&in[i]>=9999999999.99){ flag = 0; return -1; } } memset(id,-1,sizeof(id)); memset(vis,-1,sizeof(vis)); in[root] = 0; int num = 1; for(int i=1; i<=n; i++){ ans += in[i]; // debug(ans); int v = i; while(vis[v]!=i && v!=root&&id[v]==-1){ vis[v] = i; v = pre[v]; } if(v!=root&&id[v] ==-1){ for(int u=pre[v] ; u!=v; u = pre[u]){ id[u] = num; } id[v]=num++; } } if(num==1)break; for(int i=1; i<=n; i++){ if(id[i] == -1){ id[i]=num++; } } for(int i=1; i<=m; i++){ int v = e[i].v; e[i].u = id[e[i].u]; e[i].v = id[e[i].v]; if(e[i].u!=e[i].v) e[i].dis = e[i].dis - in[v]; } n = num-1; root = id[root]; } return ans; } int main(){ while(~scanf("%d%d", &n, &m)){ for(int i=1; i<=n; i++){ scanf("%d%d", &a[i].x, &a[i].y); } for(int i=1; i<=m; i++){ scanf("%d%d", &e[i].u, &e[i].v); e[i].dis = get(e[i].u, e[i].v); } flag = 1; double tmp = zhuliu(1); if(flag==0)puts("poor snoopy"); else printf("%.2f\n",tmp); } return 0; }
POJ - 3164-Command Network 最小树形图——朱刘算法
标签:cond lin main iterator txt clu gre bitset typedef
原文地址:https://www.cnblogs.com/ckxkexing/p/9394073.html