问题描述:给定平面N个点,两边相连的代价为曼哈顿距离,求这些点的最小生成树
按一般想法,prime复杂度O(n^2),Kruskal复杂度O(n^2 logn),N很大时,这复杂度要爆炸了
但是最小生成树具有一个性质——环切性质,即如果在一个图中存在一个环,把环中权最大的边删去,那么现在最小生成树的权和
删之前相同,所以很多边都是没用的,可以删去
在平面内,分割成八个区域
可以证明,中心的原点只需和每个区域的一个点相连即可(证明当然是要看莫队的)
所以构造的图中至多有8n条边,复杂度为O(nlogn)
以区域1为例,设原点为O(x0,y0),另一点P(x1,y1),则 x1>x0 且 y1-x1 > y0-y1,满足x1+y1最小的点即是最近的点
实现使先将x排序,再将y-x排序,然后用树状数组维护x+y的最小值
然后八个区域转换后可以到区域1,因为是双向边,所以四次转换就可以
第一次直接计算,第二次按y=x翻转,即x与y坐标互换,第三次按x=0翻转,即x坐标为负数,第四次在按y=x翻转,就可以计算了
7231: Spin A Web
时间限制: 1 Sec 内存限制: 128 MB
提交: 127 解决: 24
[提交] [状态] [讨论版] [命题人:admin]
题目描述
We have a canvas divided into grid with H rows and W columns. The square at the ith row from the top and the jth column from the left is represented as (i, j). (i, j) square has two value xi,j and yi,j .
Now we want to merge the squares to a connected web with minimal cost. Two squares can be connected if they are in the same row or column, and the cost of connecting (i0, j0) and (i1, j1) is
|xi0,j0 − xi1,j1 | + |yi0,j0 − yi1,j1 |.
输入
Input is given from Standard Input in the following format:
H W
x1,1 x1,2 . . . x1,W
.
.
xH,1 xH,2 . . . xH,W
y1,1 y1,2 . . . y1,W
.
.
yH,1 yH,2 . . . yH,W
Constraints
1 ≤ H × W ≤ 100000
−108 ≤ xi,j, yi,j ≤ 108(1 ≤ i ≤ H, 1 ≤ j ≤ W )
All of them are integers.
输出
Print one line denotes the minimal cost to merge the square to be a connected web.
样例输入
1 3 1 3 2 1 2 3
样例输出
5
这个题只能同行或者同列相连,所以对每行每列求一次
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 100;
const int inf = 0x3f3f3f3f;
inline int read() {
char c;
int sum = 0;
int f = 1;
c = getchar();
while (c < '0' || c > '9') {
if (c == '-')f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
sum = sum * 10 + c - '0';
c = getchar();
}
return sum * f;
}
struct node {
int x, y, pos;
friend bool operator<(node a, node b) {
if (a.x != b.x)return a.x < b.x;
return a.y < b.y;
}
} s[maxn];
int p[maxn];
struct node2 {
int u, v, val;
friend bool operator<(node2 x, node2 y) {
return x.val < y.val;
}
} edg[maxn * 8];
int tot = 0;
int a[maxn], b[maxn];
bool cmp(int a, int b) {
if (s[a].x != s[b].x) return s[a].x < s[b].x;
return s[a].y < s[b].y;
}
bool cmp2(node2 x, node2 y) {
return x.val < y.val;
}
int value[maxn], poss[maxn];
int fa[maxn];
void add(int x, int val, int pos) {
for (int i = x; i > 0; i -= (i & (-i))) {
if (value[i] > val) {
value[i] = val;
poss[i] = pos;
}
}
}
int ask(int x, int n) {
int ma = inf;
int ans = -1;
for (int i = x; i <= n; i += (i & (-i))) {
if (value[i] < ma) {
ma = value[i];
ans = poss[i];
}
}
return ans;
}
int findd(int x) {
return fa[x] == x ? x : fa[x] = findd(fa[x]);
}
void Union(int x, int y) {
int fx = findd(x);
int fy = findd(y);
if (fx != fy) {
fa[fx] = fy;
}
}
void mandis(int n) {
sort(p, p + n, cmp);
for (int i = 0; i < n; i++) {
a[i] = b[i] = s[p[i]].y - s[p[i]].x;
}
sort(b, b + n);
int k = unique(b, b + n) - b;
for (int i = 1; i <= n; i++) {
value[i] = inf;
poss[i] = -1;
}
for (int i = n - 1; i >= 0; i--) {
int pp = lower_bound(b, b + k, a[i]) - b + 1;
int ans = ask(pp, n);
if (ans != -1) {
edg[tot].u = s[p[i]].pos;
edg[tot].v = s[p[ans]].pos;
edg[tot++].val = abs(s[p[i]].x - s[p[ans]].x) + abs(s[p[i]].y - s[p[ans]].y);
}
add(pp, s[p[i]].x + s[p[i]].y, i);
}
}
void change(int n) {
for (int i = 1; i <= 4; i++) {
for (int j = 0; j < n; j++) {
if (i == 2 || i == 4)
swap(s[p[j]].x, s[p[j]].y);
else if (i == 3)
s[p[j]].x = -1 * s[p[j]].x;
}
mandis(n);
}
}
int main() {
int h, w;
h = read(), w = read();
int id = 1;
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
s[i * w + j].x = read();
s[i * w + j].pos = id++;
}
}
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
s[i * w + j].y = read();
}
}
int cnt;
for (int i = 0; i < h; i++) {
cnt = 0;
for (int j = 0; j < w; j++)
p[cnt++] = i * w + j;
change(cnt);
}
for (int j = 0; j < w; j++) {
cnt = 0;
for (int i = 0; i < h; i++) {
p[cnt++] = i * w + j;
}
change(cnt);
}
for (int i = 1; i <= h * w; i++)
fa[i] = i;
sort(edg, edg + tot, cmp2);
ll ans = 0;
for (int i = 0; i < tot; i++) {
if (findd(edg[i].u) != findd(edg[i].v)) {
Union(edg[i].u, edg[i].v);
ans += edg[i].val;
}
}
printf("%lld\n", ans);
return 0;
}