DFS
排列数字
#include<iostream>
using namespace std;
const int N = 10;
int a[N], b[N];
int n;void dfs(int u){if(u > n){for(int i = 1; i <= n; i++)cout<<a[i]<<" ";cout<<endl;return;}for(int i = 1; i <= n; i++){if(!b[i]){b[i] = 1;a[u] = i;dfs(u + 1);b[i] = 0;}}
}int main(){cin>>n;dfs(1);return 0;
}
n-皇后问题
#include<iostream>
using namespace std;
const int N = 20;
char g[N][N];
int a[N], b[N], c[N];
int n;void dfs(int u){if(u > n){for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++)cout<<g[i][j];cout<<endl;}cout<<endl;return;}for(int i = 1; i <= n; i++){if(!a[i] && !b[u + i] && !c[-u + i + n]){a[i] = b[u + i] = c[-u + i + n] = 1;g[u][i] = 'Q';dfs(u + 1);g[u][i] = '.';a[i] = b[u + i] = c[-u + i + n] = 0;}}
}int main(){cin>>n;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)g[i][j] = '.';dfs(1);return 0;
}
BFS
走迷宫
#include<iostream>
#include<cstring>
using namespace std;
const int N = 110;
int g[N][N], d[N][N];
pair<int, int> q[N * N];
int hh, tt = - 1;
int n, m;int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};void bfs(int x, int y){memset(d, -1, sizeof(d));q[++tt] = make_pair(x, y);d[x][y] = 0;while(hh <= tt){auto t = q[hh++];for(int i = 0; i < 4; i++){int a = dx[i] + t.first, b = dy[i] + t.second;if(a < 1 || a > n || b < 1 || b > m) continue;if(d[a][b] != -1) continue;if(g[a][b] != 0) continue;d[a][b] = d[t.first][t.second] + 1;q[++tt] = make_pair(a, b);}}cout<<d[n][m];
}int main(){cin>>n>>m;for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++)cin>>g[i][j];bfs(1, 1);return 0;
}
八数码
#include<iostream>
#include<unordered_map>
using namespace std;
const int N = 1e6; //一共有9!种情况
unordered_map<string, int> d;
string q[N];
int hh, tt = -1;
int n = 9;int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};int bfs(string s){q[++tt] = s;d[s] = 0;//记录终点string end = "12345678x";while(hh <= tt){string t = q[hh++];//存储当前位置到起点的距离int dis = d[t];//如果到终点了，那就返回距起点距离if(t == end) return dis;//查找x的下标int k = t.find('x');//x在矩阵中的位置int x = k / 3, y = k % 3;for(int i = 0; i < 4; i++){int a = x + dx[i], b = y + dy[i];if(a < 0 || a > 2 || b < 0 || b > 2) continue;//转移xswap(t[k], t[3 * a + b]);//如果没有遍历过，那就存储到队列中if(!d.count(t)){d[t] = dis + 1;q[++tt] = t;}//还原swap(t[k], t[3 * a + b]);}}return -1;
}int main(){char c;string s = "";for(int i = 0; i < n; i++){cin>>c;s += c;}cout<<bfs(s);return 0;
}
树和图的存储
树是一种特殊的图
存储可以用链式向前星或者vector
//链式向前星
#include<iostream>
#include<cstring>
using namespace std;
const int N = 1e5 + 10, M = 2 * N;
int h[N], e[N], ne[N], idx;
int st[N];void add(int a, int b){e[idx] = b;ne[idx] = h[a];h[a] = idx;idx++;
}void dfs(int u){st[u] = 1;for(int i = u; i != -1; i = ne[i]){int j = e[i];if(!st[j]) dfs(j);}
}int main(){memset(h, -1, sizeof(h));return 0;
}//vector存储
#include<iostream>
#include<vector>
using namespace std;
const int N = 1e5 + 10;
vector<int> v[N];
int st[N];void add(int a, int b){v[a].push_back(b);v[b].push_back(a);
}void dfs(int u){st[u] = 1;for(int i = 0; i < v[u].size(); i++){int j = v[u][i];if(!st[j]) dfs(j);}
}int main(){return 0;
}
树与图的深度优先遍历
树的重心
#include<iostream>
#include<cstring>
using namespace std;
const int N = 1e5 + 10, M = 2 * N;
int h[N], e[M], ne[M], idx;
int st[N];
int n, ans = 1e9;void add(int a, int b){e[idx] = b;ne[idx] = h[a];h[a] = idx;idx++;
}int dfs(int u){st[u] = 1;//cnt存储以u为根的节点数（包括u），res是删除掉某个节点后的最大连通子图节点数int cnt = 1, res = 0; for(int i = h[u]; i != -1; i = ne[i]){int j = e[i];if(!st[j]){//以u为节点的单棵子树的节点数int t = dfs(j);//计算以j为根的树的节点数cnt += t;//记录最大连通子图节点数res = max(res, t);}}//以u为重心，最大的连通子图节点数res = max(res, n - cnt);ans = min(ans, res);return cnt;
}int main(){memset(h, -1, sizeof(h));cin>>n;int a, b;for(int i = 0; i < n - 1; i++){cin>>a>>b;add(a, b);add(b, a);}dfs(1);cout<<ans;return 0;
}
树与图的宽度优先遍历
图中点的层次
#include<iostream>
#include<cstring>
using namespace std;
const int N = 1e5 + 10, M = 2 * N;
int h[N], e[M], ne[M], idx;
int q[N], d[N], hh, tt = -1;
int n, m;void add(int a, int b){e[idx] = b;ne[idx] = h[a];h[a] = idx;idx++;
}void bfs(int u){memset(d, -1, sizeof(d));q[++tt] = u;d[u] = 0;while(hh <= tt){//使用队头，弹出队头int t = q[hh++];for(int i = h[t]; i != -1; i = ne[i]){int j = e[i];if(d[j] == -1){//更新距离d[j] = d[t] + 1;//入队q[++tt] = j;}}}cout<<d[n];
}int main(){memset(h, -1, sizeof(h));cin>>n>>m;int x, y;while(m--){cin>>x>>y;add(x, y);}bfs(1);return 0;
}
拓扑排序
有向无环图也是拓扑图
入度：有多少条边指向自己
出度：有多少条边出去
有向图的拓扑序列
入度为0就是起点，出度为0就是终点
#include<iostream>
#include<cstring>
using namespace std;
const int N = 1e5 + 10;
int h[N], e[N], ne[N], idx;
int q[N], hh, tt = -1;
int n, m;
int r[N]; //存储入度void add(int a, int b){e[idx] = b;ne[idx] = h[a];h[a] = idx;idx++;
}void bfs(){//判断哪些点入度为0for(int i = 1; i <= n; i++)if(!r[i]) q[++tt] = i;while(hh <= tt){int t = q[hh++];for(int i = h[t]; i != -1; i = ne[i]){int j = e[i];r[j]--;if(!r[j]) q[++tt] = j;}}if(tt == n - 1){for(int i = 0; i <= tt; i++) cout<<q[i]<<" ";}else cout<<-1;
}int main(){memset(h, -1, sizeof(h));cin>>n>>m;int x, y;while(m--){cin>>x>>y;add(x, y);r[y]++;}bfs();return 0;
}
最短路
帮助理解
Dijkstra
Dijkstra求最短路 I
#include<iostream>
#include<cstring>
using namespace std;
const int N = 510;
int g[N][N], d[N], b[N];
int n, m;void dijkstra(int u){memset(d, 0x3f, sizeof(d));d[u] = 0;for(int i = 0; i < n; i++){int t = -1;for(int j = 1; j <= n; j++)if(!b[j] && (t == -1 || d[t] > d[j])) t = j;b[t] = 1;for(int j = 1; j <= n; j++)d[j] = min(d[j], d[t] + g[t][j]);}cout<<((d[n] == 0x3f3f3f3f) ? -1 : d[n]);
}int main(){memset(g, 0x3f, sizeof(g));cin>>n>>m;int x, y, z;while(m--){cin>>x>>y>>z;g[x][y] = min(g[x][y], z);}dijkstra(1);return 0;
}
Dijkstra求最短路 II
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int N = 2e5;
int h[N], e[N], ne[N], w[N], idx; //w[i]存储上个点到i的距离
int d[N], b[N];
int n, m;
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q; //小根堆，第一个元素存储距离，第二个元素存储下标void add(int x, int y, int z){e[idx] = y;w[idx] = z;ne[idx] = h[x];h[x] = idx;idx++;
}void dijkstra(int u){memset(d, 0x3f, sizeof(d));d[u] = 0;q.push(make_pair(0, 1));while(q.size()){auto t = q.top();q.pop();int x = t.first, y = t.second;if(b[y]) continue; //如果遍历过就退出b[y] = 1;for(int i = h[y]; i != -1; i = ne[i]){int j = e[i];if(d[j] > x + w[i]){d[j] = x + w[i];q.push(make_pair(d[j], j));}}}cout<<(d[n] == 0x3f3f3f3f ? -1 : d[n]);
}int main(){memset(h, -1, sizeof(h));cin>>n>>m;int x, y, z;while(m--){cin>>x>>y>>z;add(x, y, z);}dijkstra(1);return 0;
}
增加点权，求有多少条最短路
题目链接
#include<iostream>
#include<cstring>
using namespace std;
int g[505][505], dis[505], st[505];
int a[505], paths[505], teams[505];
int n, m, c1, c2;void dj(int u){teams[u] = a[u];paths[u] = 1;dis[u] = 0;for(int j = 0; j < n; j++){int t = -1;for(int i = 0; i < n; i++){if(!st[i] && (t == -1 || dis[t] > dis[i])){t = i;}}st[t] = 1;for(int i = 0; i < n; i++){if(dis[i] > dis[t] + g[t][i]){dis[i] = dis[t] + g[t][i]; paths[i] = paths[t]; //继承路径条数teams[i] = teams[t] + a[i]; //更新救援队人数}else if(dis[i] == dis[t] + g[t][i]){if(teams[i] < teams[t] + a[i]){teams[i] = teams[t] + a[i]; //选救援队人数更多的} paths[i] += paths[t]; //累加路径条数}}}
}int main(){memset(g, 0x3f, sizeof(g));cin>>n>>m>>c1>>c2;for(int i = 0; i < n; i++) cin>>a[i];while(m--){int x, y, z;cin>>x>>y>>z;g[x][y] = g[y][x] = min(g[x][y], z);}memset(dis, 0x3f, sizeof(dis));dj(c1);cout<<paths[c2]<<" "<<teams[c2];return 0;
}
增加边权，求花费最少
题目链接
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
int g[505][505], dis[505], st[505];
int cost[505][505], c[505], pre[505];
vector<int> path;
int n, m, s, d;void dj(int u){dis[u] = 0;c[u] = 0;for(int i = 0; i < n; i++){int t = -1;for(int j = 0; j < n; j++){if(!st[j] && (t == -1 || dis[t] > dis[j])){t = j;}}st[t] = 1;for(int j = 0; j < n; j++){if(dis[j] > dis[t] + g[t][j]){pre[j] = t;dis[j] = dis[t] + g[t][j];c[j] = c[t] + cost[t][j];}else if(dis[j] == dis[t] + g[t][j] && c[j] > c[t] + cost[t][j]){pre[j] = t;c[j] = c[t] + cost[t][j];}}}
}int main(){memset(g, 0x3f, sizeof(g));memset(dis, 0x3f, sizeof(dis));memset(c, 0x3f, sizeof(c));memset(cost, 0x3f, sizeof(cost));cin>>n>>m>>s>>d;while(m--){int x, y, z, h;cin>>x>>y>>z>>h;g[x][y] = g[y][x] = min(g[x][y], z);cost[x][y] = cost[y][x] = min(cost[x][y], h);}for(int i = 0; i < n; i++) pre[i] = i;dj(s);int q = d;while(q != s){path.push_back(q);q = pre[q];}path.push_back(s);int p = path.size();for(int i = p - 1; i >= 0; i--) cout<<path[i]<<" ";cout<<dis[d]<<" "<<c[d];return 0;
}
bellman-ford
有边数限制的最短路
如果负环在1到n的路径上，那就不存在最短路
#include<iostream>
#include<cstring>
using namespace std;
const int N = 510, M = 1e4 + 10;
int d[N], b[N]; //b数组备份
int n, m, k;
struct E{int x, y, z;
}e[M];void bellman_ford(int u){memset(d, 0x3f, sizeof(d));d[u] = 0;//最多k条边for(int i = 0; i < k; i++){//每次只更新一条串联路径，防止更新了多条串联路径memcpy(b, d, sizeof(d));for(int j = 0; j < m; j++){int x = e[j].x, y = e[j].y, z = e[j].z;d[y] = min(d[y], b[x] + z);}}if(d[n] > 0x3f3f3f3f / 2) cout<<"impossible";else cout<<d[n];
}int main(){cin>>n>>m>>k;int x, y, z;for(int i = 0; i < m; i++){cin>>x>>y>>z;e[i] = {x, y, z};}bellman_ford(1);return 0;
}
spfa
spfa求最短路
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int N = 1e5 + 10;
int h[N], e[N], ne[N], w[N], idx;
int dis[N], st[N];
int q[N], hh, tt = -1;
int n, m;void add(int x, int y, int z){e[idx] = y;w[idx] = z;ne[idx] = h[x];h[x] = idx;idx++;
}void spfa(int u){memset(dis, 0x3f, sizeof(dis));dis[u] = 0;q[++tt] = u;st[u] = 1;while(hh <= tt){int t = q[hh++];//有环，所以可能一个点会遍历两次st[t] = 0;for(int i = h[t]; i != -1; i = ne[i]){int j = e[i];if(dis[j] > dis[t] + w[i]){dis[j] = dis[t] + w[i];if(!st[j]){q[++tt] = j;st[j] = 1;}}}}if(dis[n] == 0x3f3f3f3f) cout<<"impossible";else cout<<dis[n];
}int main(){memset(h, -1, sizeof(h));cin>>n>>m;int x, y, z;for(int i = 0; i < m; i++){cin>>x>>y>>z;add(x, y, z);}spfa(1);return 0;
}
spfa判断负环
#include<iostream>
#include<cstring>
using namespace std;
const int N = 2e3 + 10, M = 1e4 + 10;;
int h[N], e[M], ne[M], w[M], idx;
int dis[N], st[N], cnt[N];
int q[N * N], hh, tt = -1; //有环的时候，一个元素可能会一直插入队列，所以要开N * N
int n, m;void add(int x, int y, int z){e[idx] = y;w[idx] = z;ne[idx] = h[x];h[x] = idx;idx++;
}void spfa(){//存在的负权回路，不一定从1开始for(int i = 1; i <= n; i++){q[++tt] = i;st[i] = 1;}while(hh <= tt){int t = q[hh++];//有环，所以可能一个点会遍历两次st[t] = 0;for(int i = h[t]; i != -1; i = ne[i]){int j = e[i];if(dis[j] > dis[t] + w[i]){dis[j] = dis[t] + w[i];cnt[j] = cnt[t] + 1;if(cnt[j] >= n){cout<<"Yes";return;}if(!st[j]){q[++tt] = j;st[j] = 1;}}}}cout<<"No";
}int main(){memset(h, -1, sizeof(h));cin>>n>>m;int x, y, z;for(int i = 0; i < m; i++){cin>>x>>y>>z;add(x, y, z);}spfa();return 0;
}
Floyd
Floyd求最短路
f(k, i, j) = f(k - 1, i, k) + f(k - 1, k, j);
#include<iostream>
using namespace std;
const int N = 210;
int f[N][N];
int n, m, k;void floyd(){for(int k = 1; k <= n; k++)for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
}int main(){cin>>n>>m>>k;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)if(i == j) f[i][j] = 0;else f[i][j] = 0x3f3f3f3f;int x, y, z;for(int i = 1; i <= m; i++){cin>>x>>y>>z;f[x][y] = min(f[x][y], z);}floyd();for(int i = 1; i <= k; i++){cin>>x>>y;//可能存在负权边if(f[x][y] > 0x3f3f3f3f / 2) cout<<"impossible"<<endl;else cout<<f[x][y]<<endl;}return 0;
}
最小生成树
Prim
Kruskal
二分图
