Amber <<Play with Trees Solutions>>
半徑

只需計算這些相鄰點Li,即為最優值了
在這之前,需要去除掉一些點,只保留最外層的那些,才能有上面那樣子交點

 /**//*
n<=200 , m<= 20000的無向圖
求一棵生成樹,使得任意兩點路徑中的最大值最小

這題就是求最小直徑生成樹了MSDT
有Kariv-Hakimi 算法

Amber的Play with Trees Solutions有這道題比較具體的解答
“最短路徑樹不見得是最小直徑生成樹”
absoulte center是邊上的一個點(不僅僅是圖的頂點),它也是最短最長路徑的中點
問題轉化為:
求absoulte center,使得它到其他頂點最大值(偏心距)最小

先FLoyd求出任意兩點最短路d[u,v]
absoulte center必在某條邊,枚舉所有邊u-v
假設它離u為a,設該點為r,
f(r) = max{min(d[u,x]+a, w[u,v]-a+d[v,x])},則為半徑了
至于確定a,Amber的文章上有比較具體的做法了,其實也就是有Kariv-Hakimi 算法了
覺得比較神奇的是枚舉邊,確定Absolute Center是針對變量a(r離u的距離)作目標的曲線

timus 1569是邊權為1時的
*/
#include<iostream>
#include<cstring>
#include<map>
#include<algorithm>
#include<stack>
#include<queue>
#include<cstring>
#include<cmath>
#include<string>
#include<cstdlib>
#include<vector>
#include<cstdio>
#include<set>
#include<list>
#include<numeric>
#include<cassert>
#include<sstream>
#include<ctime>
#include<bitset>
#include<functional>

using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 501;

int w[MAXN][MAXN];
int dist[MAXN][MAXN];
int n, m;
int cu, cv;
int pre[MAXN], in[MAXN];
double dp[MAXN], ans;//為了更準確點。用了double

void update(int u, int v)
  {
vector<pair<int,int> > vt, _vt;
 for (int i = 1; i <= n; i++) {
vt.push_back(make_pair(dist[u][i], dist[v][i]));
}
sort(vt.begin(), vt.end());
//按照dist[u,i]從小到大排,最后需要考慮點是dist[u,i] < dist[u,j] && dist[v,i] > dist[u,j], i < j
 for (int i = 0; i < vt.size(); i++) {
 while(!_vt.empty() && _vt.back().second <= vt[i].second) {
_vt.pop_back();
}
_vt.push_back(vt[i]);
}
int D = INF;//diameter
double a;
 if (_vt.size() == 1) {
 if (_vt[0].first < _vt[0].second) {
a = 0;
D = 2*_vt[0].first;
 } else {
a = w[u][v];
D = 2*_vt[0].second;
}
 }else {
 for (int i = 1; i < _vt.size(); i++) {
 if (D > w[u][v] + _vt[i-1].first + _vt[i].second) {
a = (w[u][v] + _vt[i].second - _vt[i-1].first) / 2.0;
D = w[u][v] + _vt[i-1].first + _vt[i].second;
}
}
}
 if (D < ans) {
cu = u;
cv = v;
ans = D;
dp[u] = a;
dp[v] = w[u][v] - a;
}
}

vector<pair<int, int> > edges;

void spfa()
  {
fill(pre+1, pre+1+n, -1);
fill(in+1, in+1+n, 0);
 for (int i = 1; i <= n; i++) {
 if (i != cu && i != cv) {
dp[i] = INF;
}
}
in[cu] = in[cv] = 1;
queue<int> q;
q.push(cu);
q.push(cv);
 while (!q.empty()) {
int t = q.front();
q.pop();
in[t] = 0;
 for (int i = 1; i <= n; i++) {
 if (w[t][i] != INF && dp[i] > dp[t] + w[t][i]) {
dp[i] = dp[t]+w[t][i];
pre[i] = t;
 if (!in[i]) {
in[i] = 1;
q.push(i);
}
}
}
}
edges.clear();
 for (int i = 1; i <= n; i++) {
 if (pre[i] != -1) {
edges.push_back(make_pair(min(i,pre[i]), max(i,pre[i])));
}
}
 if (cv != cu) {
edges.push_back(make_pair(min(cu,cv), max(cu,cv)));
}
return ;
}

int main()
  {
#ifndef ONLINE_JUDGE
freopen("in","r",stdin);
#endif
 for (;~scanf("%d%d", &n, &m); ) {
fill(w[1], w[n+1], INF);
int a, b, c;
 for (int i = 0; i < m; i++) {
scanf("%d%d%d", &a, &b, &c);
w[a][b] = w[b][a] = min(w[a][b], c);
}
 for (int i = 1; i <= n; i++) {//?
w[i][i] = 0;
}
memcpy(dist, w, sizeof dist);
for (int k = 1; k <= n; k++)
for (int i = 1; i <= n; i++)
 if (dist[i][k] != INF) {
for (int j = 1; j <= n; j++)
dist[i][j] = min(dist[i][j], dist[i][k]+dist[k][j]);
}
ans = INF;
for (int i = 1; i <= n; i++)
 for (int j = i; j <= n; j++) {
 if (w[i][j] != INF) {
update(i, j);
}
}
cout<<ans<<endl;
spfa();
sort(edges.begin(), edges.end());
 for (vector<pair<int,int> >::iterator it = edges.begin() ; it != edges.end(); it++) {
printf("%d %d\n", it->first, it->second);
}
}
return 0;
}
|
|
常用鏈接
隨筆分類
Links
搜索
最新評論

|
|