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HDU4966 GGS-DDU(最小树形图)

2024-07-18 15:57:50   219人阅读

之前几天想着补些算法的知识，学了一下最小树形图的朱刘算法，不是特别理解，备了份模板以备不时之需，想不到多校冷不丁的出了个最小树形图，没看出来只能表示对算法不太理解吧,用模板写了一下，然后就过了。－ －0

之前听到是最小树形图的时候觉得恍然大悟，非常裸，但是后来想想也不是特别裸，其实关键就是要想清楚要加回流的边，贴一份代码吧－ －0

#pragma warning(disable:4996)#include<cstdio>#include<set>#include<cstring>#include<iostream>#include<stdlib.h>#include<vector>#include<map>#include<algorithm>#include<queue>#include<cmath>#include<functional>#include<string>using namespace std;#define maxn 550int n, m;int a[55];struct Edge{	int u, v, w;	Edge(int ui, int vi, int wi) :u(ui), v(vi), w(wi){}	Edge(){}};vector<Edge> E;vector<int> vid[55];int in[maxn]; // minimum pre edge weightint pre[maxn]; // pre vertexint vis[maxn]; // vis arrayint id[maxn]; // mark down the idint nv; // nv is the number of vertex after shrinkingint directed_mst(int root,int vertex_num){	int ret = 0; int nv = vertex_num;	while (1){		for (int i = 0; i < nv; ++i) in[i] = 1e9;		for (int i = 0; i < E.size(); ++i){			int u = E[i].u, v = E[i].v;			if (E[i].w < in[v] && u != v){				in[v] = E[i].w;				pre[v] = u;			}		}		for (int i = 0; i < nv; ++i){			if (i == root) continue;			if (in[i]>1e8) return -1;		}		int cnt = 0;		memset(id, -1, sizeof(id));		memset(vis, -1, sizeof(vis));		in[root] = 0;		for (int i = 0; i < nv; ++i){			ret += in[i];			int v = i;			while (vis[v] != i&&id[v] == -1 && v != root){				vis[v] = i;				v = pre[v];			}			// v!=root means we find a circle,id[v]==-1 guarantee that it‘s not shrinked.			if (v != root&&id[v] == -1){				for (int u = pre[v]; u != v; u = pre[u]){					id[u] = cnt;				}				id[v] = cnt++;			}		}		if (cnt == 0) break;		for (int i = 0; i < nv; ++i){			if (id[i] == -1) id[i] = cnt++;		}		// change the cost of edge for each (u,v,w)->(u,v,w-in[v])		for (int i = 0; i < E.size(); ++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].w -= in[v];		}		// mark down the new root		root = id[root];		// mark down the new vertex number		nv = cnt;	}	return ret;}int main(){	while (cin >> n >> m){		if (n == 0 && m == 0) break;		int tot = 0;		for (int i = 1; i <= n; ++i) {			vid[i].clear();			scanf("%d", a + i);			for (int j = 0; j <= a[i]; ++j){				vid[i].push_back(++tot);			}		}		++tot;		E.clear();		for (int i = 1; i <= n; ++i){			for (int j = 0; j < vid[i].size(); ++j){				for (int k = j + 1; k < vid[i].size(); ++k){					E.push_back(Edge(vid[i][k], vid[i][j], 0));				}			}		}		int ci, l1, di, l2, wi;		for (int i = 0; i < m; ++i){			scanf("%d%d%d%d%d", &ci, &l1, &di, &l2, &wi);			E.push_back(Edge(vid[ci][l1], vid[di][l2], wi));		}		for (int i = 1; i <= n; ++i){			E.push_back(Edge(0, vid[i][0], 0));		}		int ans = directed_mst(0,tot);		printf("%d\n", ans);	}	return 0;}

 


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