这道题挺巧妙的,乍一看是空间上的,无从下手,稍微转换一下就可以了。
看到题目,求消除这些行星的最少次数,就是求最小割,也就是求最大流,考虑怎样建图。
考虑当我们消去一个面上的所有点时,我们消去这个面后,这个面就不会再被消了,也就是只能被消一次,比如我们消去与\(\texttt{x=1}\)垂直的面上的点后,与\(\texttt{x=1}\)垂直的这个面就不会被再消一次,\(\texttt{y,z}\)同理。
但在这个面上的某些点(\(\texttt{x}\)相同,\(\texttt{y}\),\(\texttt{z}\)不同的点)还会在另一些平面上,可能还会被再消。
直接连点的话会超时,我们考虑用点来代表面(例如\(\texttt{x}\)中的1号点就代表与\(\texttt{x=1}\)垂直的面),三个面就可以确定一个点,所以我们让\(\texttt{x}\)与\(\texttt{y}\)与\(\texttt{z}\)相连,表示这个点。
我们这样建图
建立一个超级汇点\(\texttt{s}\)和超级源点\(\texttt{t}\),\(\texttt{s}\)向所有\(\texttt{x}\)连边,\(\texttt{x}\)向\(\texttt{y}\)连边,\(\texttt{y}\)拆一下子点,拆完点出来向\(\texttt{z}\)连边,所有\(\texttt{z}\)向\(\texttt{t}\)连边。
当我们某一条\(\texttt{s->x}\)的边流满时,就代表与\(\texttt{x}\)垂直的这个面被消掉了,\(\texttt{y->y'}\)流满时表示与\(\texttt{y}\)垂直的这个面被消了,\(\texttt{z->t}\)流满时表示与\(\texttt{z}\)垂直的这个面被消了,因为我们要求最小割,所以跑一遍最大流就行了。
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int n, m, s, t, num = 1;
int head[N], cur[N], dep[N];
class node {
public :
int v, nx, w;
} e[N];
template<class T>inline void read(T &x) {
x = 0; int f = 0; char ch = getchar();
while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
x = f ? -x : x;
return ;
}
inline void add(int u, int v, int w) {
e[++num].nx = head[u], e[num].v = v, e[num].w = w, head[u] = num;
e[++num].nx = head[v], e[num].v = u, e[num].w = 0, head[v] = num;
}
queue<int>q;
bool bfs() {
memset(dep, 0, sizeof dep);
memcpy(cur, head, sizeof cur);
dep[s] = 1;
q.push(s);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = head[u]; ~i; i = e[i].nx) {
int v = e[i].v;
if (!dep[v] && e[i].w) dep[v] = dep[u] + 1, q.push(v);
}
}
return dep[t];
}
int dfs(int u, int flow) {
if (u == t) return flow;
int use = 0;
for (int &i = cur[u]; ~i; i = e[i].nx) {
int v = e[i].v;
if (e[i].w && dep[v] == dep[u] + 1) {
int di = dfs(v, min(flow, e[i].w));
e[i].w -= di, e[i ^ 1].w += di;
use += di, flow -= di;
if (flow <= 0) break;
}
}
return use;
}
int dinic() {
int ans = 0;
while (bfs()) ans += dfs(s, INF);
return ans;
}
int main() {
memset(head, -1, sizeof head);
read(n), read(m);
for (int i = 1, x, y, z; i <= n; ++i) read(x), read(y), read(z), add(x, y, z);
s = 1, t = m;
printf("%d\n", dinic());
}
原文:https://www.cnblogs.com/lykkk/p/10798180.html