图论 2023.11.27

Kruskal定义不同的优先级

P3623 [APIO2008] 免费道路
给定一个无向图，其中一些边是0，其他边为1
两个不同的点之间都应该一条且仅由一条边连接
并保持刚好K条0，求是否有解决方案
n<=2e4,m<=1e5
Kruskal定义不同的优先级
思路：Kruskal先以1边优先,筛出必须要的0边
之后Kruskal先添加必须要的的0边,再以0边优先，把0边添加至k后,再添1边，最后还要检查图的连通性
 

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e6 + 10;
int n, m, k, fa[N], tot, cnt;
struct edge {int u, v, w;
}e[N], ans[N];
bool cmp1(edge e1, edge e2) {return e1.w > e2.w;
}
bool cmp2(edge e1, edge e2) {return e1.w < e2.w;
}
int find(int x) {return x == fa[x] ? x:fa[x] = find(fa[x]);
}
void init() {cnt = tot = 0;for (int i = 1; i <= n; i++) fa[i] = i;
}
void check() {int tmp = find(1);for (int i = 2; i <= n; i++) {int f = find(i);if (f != tmp) {printf("no solution\n");exit(0);}tmp = f;}
}
int main()
{cin >> n >> m >> k;for (int i = 1; i <= m; i++)cin >> e[i].u >> e[i].v >> e[i].w;init();sort(e + 1, e + m + 1, cmp1);     //从大到小(1边优先)for (int i = 1; i <= m; i++) {int x = find(e[i].u);int y = find(e[i].v);if (x == y)  continue;fa[x] = y;if (e[i].w == 0) {tot++, e[i].w = -1;    //定为必须边，下次的Kruskal此边为最高优先级}}if (tot > k) {               //(1边优先)0边>k,0边优先时,0边依然>kprintf("no solution\n");return 0;}check();init(); sort(e + 1, e + m + 1, cmp2);for (int i = 1; i <= m; i++) {int f1 = find(e[i].u), f2 = find(e[i].v);if (f1 == f2) continue;if (e[i].w == 1 || tot < k) {fa[f1] = f2;if (e[i].w < 1) {tot++, e[i].w = 0;}ans[++cnt] = e[i];}}if (tot < k) {printf("no solution\n");return 0;}check();for (int i = 1; i <= cnt; i++) {printf("%d %d %d\n", ans[i].u, ans[i].v, ans[i].w);}return 0;
}

几何联通

P3958 [NOIP2017 提高组] 奶酪
给定一个它的高度为 h,它的长度和宽度我们可以认为是无限大的奶酪，有n个空洞坐标为（x,y,z）
统一半径为r，能否利用已有的空洞跑 到奶酪的上表面去
深度优先搜索,不需要回溯，进入和出来判断只需看z-r,z+r和0,h的比较
n<=1e3,h,r<=1e9
 

#include<iostream>
#include<algorithm>
#include<string.h>
#include<cstring>
#include<math.h>
using namespace std;
const int N = 1e3 + 10;
int n;
double h, r;
int vis[N];
int ans;
struct cir {double x, y, z;bool operator <(const cir& c)const { //按从低到高return z < c.z;                           }
}a[N];
double dist(double x1, double y1, double z1, double x2, double y2, double z2){return sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1) + (z2 - z1) * (z2 - z1));
}
void dfs(int now) {if (ans) return;if (a[now].z + r>= h) {ans = 1;return;}vis[now] = 1;for (int i = 1; i <= n; i++) {if (vis[i])  continue;if (dist(a[now].x, a[now].y, a[now].z, a[i].x, a[i].y, a[i].z) > 2 * r) continue;dfs(i);}
}
void solve() {memset(vis, 0, sizeof(vis));ans = 0;cin >> n >> h >> r;for (int i = 1; i <=n; i++) {cin >> a[i].x >> a[i].y >> a[i].z;}sort(a + 1, a + 1 + n);for (int i = 1; i <= n; i++) {if (a[i].z- r <= 0) {dfs(i);    //深搜}}if (!ans) {cout << "No" << '\n';}else {cout << "Yes" << '\n';}
}
int main()
{int t;cin >> t;while (t--)  solve();
}

二维联通

P2498[SDOI2012] 拯救小云公主
给定一个row行line列大小的矩阵，给定n个boss的位置,需要从左下角，到达右上角
找一条路径使到距离boss的最短距离最远,输出最远距离
n<=3000
思路：小数二分最远距离,boss点作为圆心, 参考奶酪
左边界或上边界通过这些洞和右边界或下边界联通时,问题无解
 

#include<iostream>
#include<cstring>
#include<math.h>
#include<queue>
using namespace std;
int dis[3001][3001];
int x[3001], y[3001];
int getdis(int x1, int y1, int x2, int y2) {return pow(x1 - x2, 2) + pow(y1 - y2, 2);
}
bool able(int d, double r) {return r * r * 4 > d;
}
int row, line, n;
queue<int>q;
bool vis[3001];
bool bfs(double r) {memset(vis, 0, sizeof(vis));while (!q.empty())q.pop();for (int i = 1; i <= n; i++) {if (row - y[i] < r||x[i]<r) {   // 左下角(row, 0)出发通过这些圆q.push(i), vis[i] = 1;}}while (!q.empty()) {int p = q.front();q.pop();if (y[p] < r|| line - x[p] < r) {      //右上角(0,line)通过这些圆return 0;}for (int i = 1; i <= n; i++)if (!vis[i] && able(dis[p][i], r)) {       vis[i] = 1, q.push(i);}}return 1;
}
int main() {cin >> n >> line >> row;line--; row--;for (int i = 1; i <= n; i++) {cin >> x[i] >> y[i];x[i]--;y[i]--;}for (int i = 1; i <= n; i++)for (int j = 1; j < i; j++)dis[i][j] = dis[j][i] = getdis(x[i], y[i], x[j], y[j]);double l = 0, r = min(row, line), mid;for (int i = 1; i <= 60; i++) {mid = (l + r) / 2;if (bfs(mid))l = mid;else r = mid;}printf("%.2lf\n", l);return 0;
}

视为T在S之间滑动，转换为拓扑图目标所有窗口匹配，即度数为0    (atcoder 329 E)

给定一个长度为n字符串S,长度为m的字符串T

将长度为n字符串全是#变成S，操作：用T替换S的部分区间

n<=2e5,m<=5

思路：视为T在S之间滑动，于是有n-m+1的窗口

转换为拓扑图：入度为每个窗口不匹配的数量，一旦为0完全匹配，边为该窗口不匹配的字符，编号为(窗口编号i相连字符编号i+j)

之后拓扑，ans倒序后得到操作的顺序，目标为所有窗口完全匹配因为（视为T在S之间滑动）

#include<bits/stdc++.h>#include<iostream>#include<algorithm>#include<map>#include<set>#include<queue>#include<cstring>#include<math.h>#include<map>#include<vector>#include<stack>#include<unordered_map>#include<unordered_set>#include<bitset>#include<array>using namespace std;#define endl '\n'#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)#define int long long#define int128 __int128_t#define ll long long#define ld long doubletypedef unsigned long long ull ;#define fr(i,l,r) for(int i=l;i<=r;i++)#define fer(i,x)   for(int i=e.head[x];i;i=e.next[i])#define ufr(i,n,z) for(int i = n;i >= z; i--)#define pb(x) push_back(x)#define all(a) a.begin(),a.end()#define fi first#define se secondtypedef pair<int,int> pr;const int N = 1e6+10;const int mod=998244353,inf=1e18;int n,m;int a[N];map<int,int>mp;void solve(){cin>>n>>m;string s,t;cin>>s>>t;vector<int>in(n+1,m);vector<vector<int>>e(n+1);            //边                        queue<int>q;for(int i=0;i<n-m+1;i++){for(int j=0;j<m;j++){if(s[i+j]==t[j]){in[i]--;if(in[i]==0){              //完全与t对应q.push(i);}}else{e[i+j].push_back(i);          }}}vector<int>ans;while(!q.empty()){int u=q.front();q.pop();ans.push_back(u);for(int i=0;i<m;i++){if(mp[u+i]){continue;}mp[u+i]=1;for(auto it:e[u+i]){in[it]--;if(in[it]==0){q.push(it);}}}}reverse(ans.begin(),ans.end());                //每一次操作位置if(ans.size()<n-m+1){cout<<"No"<<'\n';}else{cout<<"Yes"<<'\n';}}signed main(){ios;int t=1;//   cin>>t;while(t--) solve();return 0;}