题目大意
给出一个缺若干条边的树,现在让你填一些长度为定值的边,使得整个树的直径最小。
思路
给一个详细的网址,讲的非常明白。
http://www.ccf.org.cn/resources/1190201776262/fujian/xuhaoran2013-07-25-03_33_55.pdf
还有数据范围是50w。
CODE
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 500010
using namespace std;
int points,edges,cost;
int head[MAX],total;
int _next[MAX << 1],aim[MAX << 1],length[MAX << 1];
int father[MAX],dis[MAX];
int f[MAX];
int Find(int x)
{
if(f[x] == x) return x;
return f[x] = Find(f[x]);
}
inline void Add(int x,int y,int len)
{
_next[++total] = head[x];
aim[total] = y;
length[total] = len;
head[x] = total;
}
int max_length[MAX],next_length[MAX];
int max_son[MAX];
void DFS(int x,int last)
{
father[x] = last;
for(int i = head[x]; i; i = _next[i]) {
if(aim[i] == last) continue;
DFS(aim[i],x);
if(max_length[aim[i]] + length[i] > max_length[x]) {
max_length[x] = max_length[aim[i]] + length[i];
max_son[x] = aim[i];
}
}
for(int i = head[x]; i; i = _next[i]) {
if(aim[i] != last && aim[i] != max_son[x])
next_length[x] = max(next_length[x],max_length[aim[i]] + length[i]);
}
}
int g[MAX];
int cnt,now_tree;
int __max[MAX],__min[MAX];
void Calc(int x,int last)
{
for(int i = head[x]; i; i = _next[i]) {
if(aim[i] == last) continue;
g[aim[i]] = max(g[x],max_son[x] == aim[i] ? next_length[x]:max_length[x]) + length[i];
int now = max(g[aim[i]],max_length[aim[i]]);
__max[now_tree] = max(__max[now_tree],now);
__min[now_tree] = min(__min[now_tree],now);
Calc(aim[i],x);
}
}
int main()
{
cin >> points >> edges >> cost;
for(int i = 1; i <= points; ++i)
f[i] = i;
for(int x,y,z,i = 1; i <= edges; ++i) {
scanf("%d%d%d",&x,&y,&z);
++x,++y;
Add(x,y,z),Add(y,x,z);
int fx = Find(x),fy = Find(y);
if(fx != fy)
f[fx] = fy;
}
int ans = 0;
int max_1 = 0,max_2 = 0,max_3 = 0;
for(int i = 1; i <= points; ++i)
if(f[i] == i) {
++cnt;
now_tree = i;
DFS(i,0);
g[i] = 0;
__max[i] = __min[i] = max_length[i];
Calc(i,0);
ans = max(ans,__max[i]);
if(__min[i] > max_1) {
max_3 = max_2;
max_2 = max_1;
max_1 = __min[i];
}
else if(__min[i] > max_2) {
max_3 = max_2;
max_2 = __min[i];
}
else
max_3 = max(max_3,__min[i]);
}
if(cnt > 1) ans = max(ans,max_1 + max_2 + cost);
if(cnt > 2) ans = max(ans,max_2 + max_3 + (cost << 1));
cout << ans << endl;
return 0;
}
时间: 2024-10-28 05:40:58