题意:求以1为根的最小树形图 没有输出字符串
思路:直接高朱刘算法 不懂的可以百度 学会了就是直接套模板的事情 其实就是不断消圈而已 不构成圈就有解 无法从根到达其他点就无解
#include <cstdio> #include <cstring> #include <cmath> const int maxn = 110; const int maxm = 50010; const double INF = 999999999; struct edge { int u, v; double w; edge(){} edge(int u, int v, double w) : u(u), v(v), w(w){} }e[maxm]; struct Point { double x, y; }p[maxn]; int pre[maxn], vis[maxn], no[maxn]; double in[maxn]; double dis(Point a, Point b) { return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); } double MST(int n, int m, int rt) { double ans = 0; while(1) { for(int i = 1; i <= n; i++) in[i] = INF; for(int i = 1; i <= m; i++) { int u = e[i].u; int v = e[i].v; if(u != v && in[v] > e[i].w) { pre[v] = u; in[v] = e[i].w; } } for(int i = 1; i <= n; i++) { if(i == rt) continue; if(in[i] == INF) return -1; } memset(no, -1, sizeof(no)); memset(vis, -1, sizeof(vis)); in[rt] = 0; int cnt = 0; for(int i = 1; i <= n; i++) { ans += in[i]; int v = i; while(v != rt && vis[v] != i && no[v] == -1) { vis[v] = i; v = pre[v]; } if(v != rt && no[v] == -1) { for(int u = pre[v]; u != v; u = pre[u]) no[u] = cnt+1; no[v] = cnt+1; cnt++; } } if(!cnt) return ans; for(int i = 1; i <= n; i++) if(no[i] == -1) no[i] = ++cnt; for(int i = 1; i <= m; i++) { int u = e[i].u; int v = e[i].v; e[i].u = no[u]; e[i].v = no[v]; if(no[u] != no[v]) { e[i].w -= in[v]; } } n = cnt; rt = no[rt]; } return ans; } int main() { int n, m; while(scanf("%d %d", &n, &m) != EOF) { for(int i = 1 ; i <= n; i++) scanf("%lf %lf", &p[i].x, &p[i].y); int add = 1; for(int i = 1; i <= m; i++) { scanf("%d %d", &e[add].u, &e[add].v); if(e[add].u == e[add].v) continue; e[add].w = dis(p[e[add].u], p[e[add].v]); add++; } double ans = MST(n, add-1, 1); if(ans < 0) puts("poor snoopy"); else printf("%.2f\n", ans); } return 0; }
POJ 3164 Command Network 最小树形图-朱刘算法裸题,布布扣,bubuko.com
POJ 3164 Command Network 最小树形图-朱刘算法裸题
原文地址:http://blog.csdn.net/u011686226/article/details/26745453