「网络流 24 题」魔术球
题解:以样例中的答案11作解释,1-11之间可以连接一些边,比如1-3,1-8等等,我们从源点向Xi连边,从Yi向汇点连边,将连接的边变成Xu到Yv,那么我们的问题就变成了最小路径覆盖为n的图中最多允许存在多少个点?
题解:这样我们从小打到一次次尝试加点,同时根据网络流的性质,我们可以直接在残量网络上跑二分图匹配,然后问题就和最小路径覆盖一模一样了。
最小路径覆盖
#include <bits/stdc++.h>
const int MAXN = 100000,INF = 10000000;
using namespace std;
int n,e = 2,ans,mf,s,t;
int d[MAXN],cur[MAXN],head[MAXN];
bool vis[MAXN];
struct node{
int v,c,f,next;
}edge[MAXN];
inline void addedge(int u,int v,int w){
edge[e] = (node){v,w,0,head[u]};head[u] = e++;
edge[e] = (node){u,0,0,head[v]};head[v] = e++;
}
inline bool check(int u,int v){
double sum = u+v;
double sq = sqrt(sum);
if(sum / sq == sq && sq == (int)sq) return true;
else return false;
}
inline void init(){
s = 0;t = 4000;
ans = n;
for(int i = 1;i <= ans;i++) addedge(s,i,1);
for(int i = 1;i <= ans;i++) addedge(i+2000,t,1);
for(int i = 1;i <= ans;i++)
for(int j = i + 1;j <= ans;j++)
if(check(i,j))
addedge(i,j+2000,1);
}
queue<int>Q;
inline bool bfs(){
memset(d,0,sizeof(d));
Q.push(s);d[s] = 1;
while(!Q.empty()){
int x = Q.front();Q.pop();
for(int i = head[x];i;i = edge[i].next){
node E = edge[i];
if(!d[E.v] && E.c > E.f){
d[E.v] = d[x] + 1;
Q.push(E.v);
}
}
}
return d[t] ? true : false;
}
int dfs(int x,int a){
if(x == t || a == 0) return a;
int flow = 0,f;
for(int &i = cur[x];i;i = edge[i].next){
node &E = edge[i];
if(d[E.v] == d[x] + 1 && (f = dfs(E.v,min(E.c - E.f,a))) > 0){
E.f += f;
edge[i^1].f += f;
a -= f;
flow += f;
if(a == 0) break;
}
}
return flow;
}
inline int max_flow()
{
int flow = 0;
while(bfs()){
for(int i = s;i <= t;i++) cur[i] = head[i];
flow += dfs(s,INF);
}
return flow;
}
void cut(int x){
vis[x] = true;printf("%d ",x);
for(int i = head[x];i;i = edge[i].next){
node E = edge[i];
if(1 <= E.v - 2000&& E.v - 2000 <= ans&& !vis[E.v] && E.c == E.f){
cut(E.v-2000);
}
}
}
int main()
{
scanf("%d",&n);
init();mf = max_flow();
while(ans - mf <= n){
ans++;
addedge(s,ans,1);
addedge(ans+2000,t,1);
for(int i = 1;i < ans;i++) if(check(i,ans)) addedge(i,ans+2000,1);
mf += max_flow();
}
ans--;
printf("%d\n",ans);
for(int i = 1;i <= ans;i++){
if(!vis[i]){
cut(i);
printf("\n");
}
}
return 0;
}