BZOJ2001 [Hnoi2010]City 城市建設 【CDQ分治 + kruskal】
阿新 • • 發佈:2018-05-21
city 刪掉 fin www 題目 ref air pan PE 了
復雜度我也不知道是什麽
題目鏈接
BZOJ2001
題解
CDQ分治神題。。。
難想難寫。。
比較樸素的思想是對於每個詢問都求一遍\(BST\),這樣做顯然會爆
考慮一下時間都浪費在了什麽地方
我們每次求\(BST\)實際上就只有一條邊不同,我們實際浪費了很多時間在處理相同的邊上
那就考慮分治
對於一個待修改的邊集,我們將其權值全部設為\(-\infty\),跑一遍\(BST\),此時其它邊如果被選中,說明這些邊在單獨詢問時也一定會被選,將這些邊連的點縮點
同樣,對於一個待修改的邊集,我們將其權值全部設為\(\infty\),跑一遍\(BST\),此時其它邊沒被選中,說明這些邊在單獨詢問時也一定不會被選,將這些邊刪掉
這樣就可以\(A\)
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 50005,maxm = 50005,INF = 0x3fffffff;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
struct EDGE{int i,a,b,w;}e[50][maxm],d[maxm],b[maxm];
struct Que{int u,v;}q[maxm];
inline bool operator <(const EDGE& a,const EDGE& b){
return a.w < b.w;
}
LL ans[maxm];
int n,m,Q,sum[50],w[maxm],id[maxm],a[maxm];
int pre[maxm];
int find(int u){return u == pre[u] ? u : pre[u] = find(pre[u]);}
void clear(int t){
for (int i = 1; i <= t; i++){
pre[d[i].a] = d[i].a;
pre[d[i].b] = d[i].b;
}
}
void comb(int& t,LL& Ans){
clear(t); int tmp = 0,fa,fb;
sort(d + 1,d + 1 + t);
REP(i,t){
fa = find(d[i].a); fb = find(d[i].b);
if (fa != fb){
pre[fb] = fa;
b[++tmp] = d[i];
}
}
REP(i,tmp) {
pre[b[i].a] = b[i].a;
pre[b[i].b] = b[i].b;
}
REP(i,tmp) if (b[i].w != -INF){
fa = find(b[i].a); fb = find(b[i].b);
if (fa != fb){
pre[fb] = fa;
Ans += b[i].w;
}
}
tmp = 0;
REP(i,t){
fa = find(d[i].a); fb = find(d[i].b);
if (fa != fb){
b[++tmp] = d[i];
id[d[i].i] = tmp;
b[tmp].a = find(b[tmp].a);
b[tmp].b = find(b[tmp].b);
}
}
REP(i,tmp) d[i] = b[i];
t = tmp;
}
void rd(int& t){
clear(t); int tmp = 0,fa,fb;
sort(d + 1,d + 1 + t);
REP(i,t){
fa = find(d[i].a); fb = find(d[i].b);
if (fa != fb){
pre[fb] = fa;
b[++tmp] = d[i];
}
else if (d[i].w == INF){
b[++tmp] = d[i];
}
}
for (int i = 1; i <= tmp; i++) d[i] = b[i];
t = tmp;
}
void solve(int l,int r,int now,LL Ans){
int t = sum[now];
if (l == r) a[q[l].u] = q[l].v; //原標號邊權值
for (int i = 1; i <= t; i++)
e[now][i].w = a[e[now][i].i]; //邊賦值
for (int i = 1; i <= t; i++)
d[i] = e[now][i],id[d[i].i] = i; //新邊對應舊邊位置
if (l == r){
ans[l] = Ans; clear(t);
sort(d + 1,d + 1 + t);
int fa,fb;
for (int i = 1; i <= t; i++){
fa = find(d[i].a); fb = find(d[i].b);
if (fa != fb){
pre[fb] = fa; ans[l] += d[i].w;
}
}
return;
}
for (int i = l; i <= r; i++) d[id[q[i].u]].w = -INF;
comb(t,Ans);
for (int i = l; i <= r; i++) d[id[q[i].u]].w = INF;
rd(t);
REP(i,t) e[now + 1][i] = d[i];
sum[now + 1] = t;
int mid = l + r >> 1;
solve(l,mid,now + 1,Ans);
solve(mid + 1,r,now + 1,Ans);
}
int main(){
n = read(); m = read(); Q = read();
for (int i = 1; i <= m; i++){
e[0][i].i = i;
e[0][i].a = read();
e[0][i].b = read();
a[i] = e[0][i].w = read();
}
for (int i = 1; i <= Q; i++){
q[i].u = read();
q[i].v = read();
}
sum[0] = m;
solve(1,Q,0,0);
for (int i = 1; i <= Q; i++)
printf("%lld\n",ans[i]);
return 0;
}
BZOJ2001 [Hnoi2010]City 城市建設 【CDQ分治 + kruskal】