poj 3164 Command Network

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题目大意：给你n个点，m条可选的有向边，要你选择n-1条边使得一号点可以到达所有点且边权最小，求最小的边权是多少 n<=100 m<=10000

这题就是一个裸的最小树形图（朱刘算法），不多说了，直接贴代码

#include<cmath>#include<cstdio>#include<cstring>#include<cassert>#include<iostream>#include<algorithm>#define sqr(x) ((x)*(x))using namespace std;const int maxn=110,maxm=10011;const double INF=1e9;struct Tedge{int x,y; double w;}e[maxm];int n,m; double x[maxn],y[maxn];double dis(int a,int b){return sqrt(sqr(x[a]-x[b])+sqr(y[a]-y[b]));}void init(){for (int i=1;i<=n;++i) scanf("%lf%lf",x+i,y+i);for (int i=1;i<=m;++i){scanf("%d%d",&e[i].x,&e[i].y); e[i].w=dis(e[i].x,e[i].y);if (e[i].x==e[i].y) --i,--m;}}double in[maxn]; int pre[maxn],vis[maxn],be[maxn];void work(){int V=n,E=m,root=1; double ans=0;while (1){for (int i=1;i<=V;++i) in[i]=INF,pre[i]=0;for (int i=1;i<=E;++i)if (e[i].x!=e[i].y && in[e[i].y]>e[i].w) in[e[i].y]=e[i].w,pre[e[i].y]=e[i].x;for (int i=1;i<=V;++i) if (i!=root && !pre[i]) return puts("poor snoopy"),void();int cnt=0; for (int i=1;i<=V;++i) vis[i]=be[i]=0; in[root]=0;for (int i=1;i<=V;++i){ans+=in[i]; int tmp=i,t;while (vis[tmp]!=i && !be[tmp] && tmp!=root) vis[tmp]=i,tmp=pre[tmp];if (tmp!=root && !be[tmp]) for (be[tmp]=++cnt,t=pre[tmp];t!=tmp;t=pre[t]) be[t]=cnt;}if (!cnt) return printf("%.2f\n",ans),void();for (int i=1;i<=V;++i) if (!be[i]) be[i]=++cnt;for (int i=1;i<=E;++i){int x=e[i].x,y=e[i].y;if ((e[i].x=be[x])!=(e[i].y=be[y])) e[i].w-=in[y];}root=be[root]; V=cnt;}}int main(){while (scanf("%d%d",&n,&m)!=EOF) init(),work(); return 0;}

。。。上面的那个是有一定的问题的，可能会被卡到O(n^3+nm)的，下面再给出一个修改过后是严格O(n^2+nm)的

#include<cmath>#include<cstdio>#include<cstring>#include<cassert>#include<iostream>#include<algorithm>#define sqr(x) ((x)*(x))using namespace std;const int maxn=110,maxm=10011,INF=(int)1e9;struct Tedge{int x,y; double w;}e[maxm];int n,m; double x[maxn],y[maxn];double dis(int a,int b){return sqrt(sqr(x[a]-x[b])+sqr(y[a]-y[b]));}void init(){for (int i=1;i<=n;++i) scanf("%lf%lf",x+i,y+i);for (int i=1;i<=m;++i){scanf("%d%d",&e[i].x,&e[i].y); e[i].w=dis(e[i].x,e[i].y);if (e[i].x==e[i].y) --i,--m;}}double in[maxn];int pre[maxn],be[maxn],vis[maxn];void work(){int V=n,E=m,root=1; double ans=0;while (1){for (int i=1;i<=V;++i) in[i]=INF; in[root]=0; int cnt=0,ne=0;for (int i=1;i<=E;++i)if (e[i].y!=root && in[e[i].y]>e[i].w) in[e[i].y]=e[i].w,pre[e[i].y]=e[i].x;for (int i=1;i<=V;++i) if (in[i]==INF) return puts("poor snoopy"),void();for (int i=1;i<=V;++i) be[i]=vis[i]=0; vis[root]=V+1;for (int i=1;i<=V;++i){ans+=in[i]; int tmp=i,t; while (!vis[tmp]) vis[tmp]=i,tmp=pre[tmp];if (vis[tmp]==i) for (be[tmp]=++cnt,t=pre[tmp];t!=tmp;t=pre[t]) be[t]=cnt;}if (!cnt) return printf("%.2f\n",ans),void();for (int i=1;i<=V;++i) if (!be[i]) be[i]=++cnt;for (int i=1;i<=E;++i){int x=e[i].x,y=e[i].y;if (be[x]!=be[y]) e[++ne].x=be[x],e[ne].y=be[y],e[ne].w=e[i].w-in[y];}E=ne; V=cnt; root=be[root];}}int main(){while (scanf("%d%d",&n,&m)!=EOF) init(),work(); return 0;}

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