1.最短路
图全为正权使用Dijkstra，有负权用SPFA，Bellman-Ford稍加了解即可
void spfa(){ queue<int> q; for(int i = 1;i <= n;i++) d[i] = 0x7fffffff; q.push(s);vis[s] = 1;d[s] = 0; while(!q.empty()){ int x = q.front();q.pop();vis[x] = 0; for(int i = head[x];i;i = G[i].pre){ int v = G[i].to,w = G[i].v; if(d[v] > d[x] + w){ d[v] = d[x] + w; if(!vis[v]){ vis[v] = 1; q.push(v); } } } } }
struct HeapNode{ int u,d; bool operator < (const HeapNode& rhs) const{ return d > rhs.d; } }; void Dijkstra() { priority_queue<HeapNode> q; for(int i = 1;i <= n;i++) d[i] = INF; d[s] = 0; q.push((HeapNode){s,d[s]}); while(!q.empty()){ HeapNode x = q.top();q.pop(); int u = x.u; if(x.d > d[u]) continue; for(register int i = last[u];i >= 0;i = e[i].next) { int v = e[i].v,w = e[i].w; if(d[u] + w < d[v]){ d[v] = d[u] + w; q.push((HeapNode){v,d[v]}); } } } }
/* Bellman-Ford算法 题解上的 */ #include<iostream> using namespace std; const int maxx=10001; int n,m,s,dis[maxx],w[500001],num[maxx],f[maxx][maxx/10][2],a=0; int main(){ ios::sync_with_stdio(false); cin>>n>>m>>s; for(int i=1;i<=n;i++) dis[i]=400; for(int i=1;i<=m;i++) for(int j=1;j<=m;j++) w[i]=400; for(int i=1;i<=m;i++){ int x,y,v; cin>>x>>y>>v; f[x][++num[x]][0]=y; f[x][num[x]][1]=i; w[i]=v; } dis[s]=0; while(a<=50){ //循环大法好 for(int i=1;i<=n;i++) for(int j=1;j<=num[i];j++) dis[f[i][j][0]]=min(dis[f[i][j][0]],dis[i]+w[f[i][j][1]]); a++; } for(int i=1;i<=n;i++) if(dis[i]==400) cout<<2147483647<<' '; else cout<<dis[i]<<' '; return 0; }
2.最小生成树
主要掌握Kruskal算法，Prim算法稍加了解即可
/* 适用于稀疏图 */ #include <bits/stdc++.h> using namespace std; int p[5005]; int n,m,num = 0; struct node { int x,y,z; }e[200005]; int cmp(node a,node b) { return a.z < b.z; } int find(int x) { return p[x] == x ? x : p[x] = find(p[x]); } int Kruskal() { for(int i = 1;i <= n;i++) p[i] = i; sort(e+1,e+m+1,cmp); int ans = 0,cnt = 0; for(int i = 1;i <= m;i++) { int x = find(e[i].x); int y = find(e[i].y); if(x != y) { p[x] = y; ans += e[i].z; cnt++; } if(cnt == n-1) break; } return ans; } int main() { scanf("%d%d",&n,&m); for(int i = 1; i <= m;i++) { scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].z); } printf("%d ",Kruskal()); return 0; }
/* 适用于稠密图（然而没啥用） */ #include <bits/stdc++.h> using namespace std; const int maxn = 5010; const int INF = 0x7fffffff; int n,m,cnt,head[maxn],dis[maxn],vis[maxn]; struct node{ int to,v,pre; }e[400010]; void addedge(int from,int to,int v){ e[++cnt].pre=head[from]; e[cnt].to=to; e[cnt].v=v; head[from]=cnt; } int Prim(){ memset(dis,0x3f,sizeof(dis)); dis[1]=0;int ans = 0; for(int i = 1;i <= n;i++){ int minn = INF,k = 0; for(int j = 1;j <= n;j++){ if(!vis[j] && dis[j] < minn){ minn = dis[j]; k = j; } } if(minn == INF)break; vis[k] = 1; for(int j = head[k];j;j = e[j].pre){ int f = e[j].to; if(!vis[f]) dis[f] = min(dis[f],e[j].v); } } for(int i = 1;i <= n;i++) ans += dis[i]; return ans; } int main(){ scanf("%d%d",&n,&m); int x,y,z; for(int i = 1;i <= m;i++){ scanf("%d%d%d",&x,&y,&z); addedge(x,y,z);addedge(y,x,z); } printf("%d ",Prim()); return 0; }
3.LCA
主要理解倍增法，树剖也可以了解，Tarjan就算了
/* 倍增写法 常数略大 我觉得不太好理解 所以我不用倍增了 */ #include <bits/stdc++.h> using namespace std; const int MAXN = 500010; int deep[MAXN],f[MAXN][25],lg[MAXN],head[MAXN],cnt; int n,m,s; struct node{ int to,pre; }G[MAXN*2]; void add(int from,int to){ G[++cnt].to = to; G[cnt].pre = head[from]; head[from] = cnt; } inline int read() { int x = 0,m = 1; char ch; while(ch < '0' || ch > '9') {if(ch == '-') m = -1;ch = getchar();} while(ch >= '0' && ch <= '9'){x = x*10+ch-'0';ch=getchar();} return m * x; } inline void dfs(int u) { for(int i = head[u];i;i = G[i].pre) { int v = G[i].to; if(v != f[u][0]) { f[v][0] = u; deep[v] = deep[u] + 1; dfs(v); } } } inline int lca(int u,int v) { if(deep[u] < deep[v]) swap(u,v); int dis = deep[u] - deep[v]; for(register int i = 0;i <= lg[n];i++) { if((1 << i) & dis) u = f[u][i]; } if(u == v) return u; for(register int i = lg[deep[u]];i >= 0;i--) { if(f[u][i] != f[v][i]) { u = f[u][i];v = f[v][i]; } } return f[u][0]; } inline void init() { for(register int j = 1;j <= lg[n];j++) { for(register int i = 1;i <= n;i++) { if(f[i][j-1] != -1) f[i][j] = f[f[i][j-1]][j-1]; } } } int main() { int x,y,a,b; n = read();m = read();s = read(); for(register int i = 1;i <= n;i++) { lg[i] = lg[i-1] + (1 << lg[i-1] + 1 == i); } for(register int i = 1;i <= n-1;i++) { x = read();y = read(); add(x,y);add(y,x); } dfs(s); init(); while(m--) { a = read();b = read(); printf("%d ",lca(a,b)); } return 0; }
/* 这种树剖写法好理解（好背） 虽然代码比倍增略长 但是也比倍增快 简直完美2333333 所以以后就用这种方法辣 */ #include <bits/stdc++.h> using namespace std; const int maxn = 500005; int fa[maxn],top[maxn],id[maxn],son[maxn],depth[maxn],size[maxn];//树剖要用的所有数组 int n,m,s,head[maxn],cnt; struct node{ int to,pre; }G[maxn*2]; void addedge(int from,int to){ G[++cnt].to = to; G[cnt].pre = head[from]; head[from] = cnt; } void dfs1(int x){ size[x] = 1; for(int i = head[x];i;i = G[i].pre){ int cur = G[i].to; if(cur == fa[x]) continue; depth[cur] = depth[x] + 1; fa[cur] = x; dfs1(cur); size[x] += size[cur]; if(size[cur] > size[son[x]]) son[x] = cur; } } void dfs2(int x,int t){ top[x] = t; if(son[x]) dfs2(son[x],t); for(int i = head[x];i;i = G[i].pre){ int cur = G[i].to; if(cur != fa[x] && cur != son[x]) dfs2(cur,cur); } } int lca(int x,int y){ while(top[x] != top[y]){ if(depth[top[x]] < depth[top[y]]) swap(x,y); x = fa[top[x]]; } if(depth[x] > depth[y]) swap(x,y); return x; } int main(){ int x,y,a,b; scanf("%d%d%d",&n,&m,&s); for(int i = 1;i < n;i++){ scanf("%d%d",&x,&y); addedge(x,y);addedge(y,x); } dfs1(s); dfs2(s,s); while(m--){ scanf("%d%d",&a,&b); printf("%d ",lca(a,b)); } return 0; }
/* Tarjan算法（题解） 常数挺大的，不推荐，容易被卡 */ #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <cctype> #include <cstring> #include <queue> #include <map> #define ll long long #define ri register int #define ull unsigned long long using namespace std; const int maxn=500005; const int inf=0x7fffffff; template <class T>inline void read(T &x){ x=0;int ne=0;char c; while(!isdigit(c=getchar()))ne=c=='-'; x=c-48; while(isdigit(c=getchar()))x=(x<<3)+(x<<1)+c-48; x=ne?-x:x; return ; } int n,m,s,t; struct Edge{ int ne,to; }edge[maxn<<1]; struct QU{ int d,id; QU(int x,int y){d=x,id=y;} QU(){;} }; vector <QU>q[maxn]; int h[maxn],num_edge=0,ans[maxn]; inline void add_edge(int f,int to){ edge[++num_edge].ne=h[f]; edge[num_edge].to=to; h[f]=num_edge; return; } int fa[maxn],vis[maxn]; int get(int x){ if(fa[x]!=x)fa[x]=get(fa[x]); return fa[x]; } void dfs(int cur){ int u,v; vis[cur]=1; for(ri i=h[cur];i;i=edge[i].ne){ v=edge[i].to; if(vis[v])continue; dfs(v); fa[v]=cur;//dfs后再合并 } for(ri i=0;i<q[cur].size();i++){ u=q[cur][i].d,v=q[cur][i].id; if(vis[u]==2){ ans[v]=get(u); } } vis[cur]=2;//dfs过 return ; } int main(){ int x,y; read(n),read(m),read(s); for(ri i=1;i<n;i++){ read(x),read(y); add_edge(x,y); add_edge(y,x); fa[i]=i; }fa[n]=n; for(ri i=1;i<=m;i++){ read(x),read(y); //q[x].push_back(y);q[y].push_back(x); q[x].push_back(QU(y,i)); q[y].push_back(QU(x,i)); } dfs(s); for(ri i=1;i<=m;i++){ printf("%d ",ans[i]); } return 0; }
4.二分图最大匹配
要会
#include <bits/stdc++.h> using namespace std; const int maxn = 1e3+5; vector<int> G[maxn]; int link[maxn],vis[maxn],n,m,e; bool dfs(int x){ for(int i = 0;i < G[x].size();i++){ int v = G[x][i]; if(!vis[v]){ vis[v] = 1; if(!link[v] || dfs(link[v])){ link[v] = x;return true; } } } return false; } int main() { int x,y; scanf("%d%d%d",&n,&m,&e); for(int i = 1;i <= e;i++){ scanf("%d%d",&x,&y); if(x <= n && y <= m) G[x].push_back(y); } int ans = 0; for(int i = 1;i <= n;i++){ memset(vis,0,sizeof(vis)); if(dfs(i)) ans++; } printf("%d ",ans); return 0; }
5.Tarjan
全部要会
vector<int> G[maxn]; //vector邻接表存图 int pre[maxn],low[maxn],sccno[maxn],cnt,scccnt; //pre[i]表示节点i被搜到的次序，lowlink[i]表示i及其后代能追溯到的最早的点v的 //pre[v]值，sccno[i]就是i所在的强连通分量的编号，dfs_clock表示第几次dfs， //scc_cnt表示找到的强连通分量序号的临时值 stack<int> S;//存储dfs到的每一个点 void dfs(int u) { pre[u] = low[u] = ++cnt;//先把pre和lowlink初始化为dfs的时间戳 S.push(u); for(int i = 0;i < G[u].size();i++)//遍历与点u相连的所有点 { int v = G[u][i];//取点 if(!pre[v]){//如果点v没有遍历过 dfs(v);//深搜 low[u] = min(low[u],low[v]);//向上合并lowlink的值 } else if(!sccno[v])//此时v已经搜过，但是不属于任何一个scc，那么就说明已经形成了环 { low[u] = min(low[u],pre[v]); } } if(low[u] == pre[u]){//如果u为最先搜到的点，它就是这个scc的根节点 scccnt++; for(;;) { int x = S.top();S.pop();//取一个搜过的点 sccno[x] = scccnt;//它属于这个强联通分量 if(x == u) break;//直到栈中 } } } void find_scc(int n) { cnt = scccnt = 0; memset(sccno,0,sizeof(sccno)); memset(pre,0,sizeof(pre)); for(int i = 1;i <= n;i++) if(!pre[i]) dfs(i); }
/* 易得出状态转移为val[v] = max(val[v],val[pre] + a[v]) 于是通过tarjan算法把所有环缩为一点,跑一遍DAG上的DP即可 */ #include <bits/stdc++.h> using namespace std; const int maxn = 10e5+5; int scc[maxn],dfn[maxn],low[maxn],stac[maxn]; int a[maxn],head1[maxn],head2[maxn],vis[maxn]; int val[maxn],cnt,cnt1,scc_clock,top,n,m,tot,scc_amount; struct node{ int to,from,pre; }g1[maxn*2],g2[maxn*2];//g1为原来的图，g2为缩点后的图 void add1(int from,int to){ g1[++cnt].from = from; g1[cnt].to = to; g1[cnt].pre = head1[from]; head1[from] = cnt; } //对应g1的add操作 void add2(int from,int to){ g2[++cnt1].from = from; g2[cnt1].to = to; g2[cnt1].pre = head2[from]; head2[from] = cnt1; } //对应g2的add操作 void tarjan(int x){ low[x] = dfn[x] = ++scc_clock;//初始化为时间戳 stac[++top] = x;vis[x] = 1;//入栈、标志数组设为1 for(register int i = head1[x];i;i = g1[i].pre){//遍历与x连接的点 int v = g1[i].to; if(!dfn[v]){ tarjan(v); low[x] = min(low[x],low[v]); }else if(vis[v]){ low[x] = min(low[x],dfn[v]); } }//一顿tarjan的操作 if(low[x] == dfn[x]){ scc_amount++; int y; while(y = stac[top--]){//出栈 scc[y] = x; vis[y] = 0;//我也不知道为啥 if(x == y) break; a[x] += a[y];//加权值 } } } int dfs(int u){ val[u] = a[u]; for(int i = head2[u];i;i = g2[i].pre){ int v = g2[i].to; dfs(v); val[v] = max(val[v],val[u] + a[v]); } } int main(){ int x,y; scanf("%d%d",&n,&m); for(register int i = 1;i <= n;i++){ scanf("%d",a+i); } for(register int i = 1;i <= m;i++){ scanf("%d%d",&x,&y); add1(x,y); } for(register int i = 1;i <= n;i++){ if(!dfn[i]) tarjan(i);//tarjan算法基本操作 } for(register int i = 1;i <= m;i++){ x = scc[g1[i].from],y = scc[g1[i].to]; if(x != y){//如果不属于同一强连通分量，就连边 add2(x,y); } } int ans = 0; for(int i = 1;i <= scc_amount;i++){ if(!val[i]){ dfs(i); ans = max(ans,val[i]); } } printf("%d ",ans); return 0; }
/* 无向图割点 对该图进行一次 Tarjan 算法（这里注意在搜索树中把无向边当做有向边看。即LOW[u]=min(LOW[u],DFN[v])(v 是 u 的祖先)的条件变为(v 是 u 的祖先且 v 不是 u 的父亲)）这样之后枚举搜索树上的所有边(u,v)，若存在 LOW[v]>=DNF[u]，则 u 是割点。 无向图割边 对该图进行一次 Tarjan 算法（这里注意在搜索树中把无向边当做有向边看。即LOW[u]=min(LOW[u],DFN[v])(v 是 u 的祖先)的条件变为(v 是 u 的祖先且 v 不是 u 的父亲)）这样之后枚举搜索树上的所有边(u,v)，若存在 LOW[v]>DNF[u]，则(u,v)为割边。 */ #include <bits/stdc++.h> using namespace std; const int maxn = 100005; int low[maxn],dfn[maxn],cut[maxn]; int n,m,cnt; vector<int> G[maxn]; void tarjan(int x,int father){ int child = 0; low[x] = dfn[x] = ++cnt; for(int i = 0;i < G[x].size();i++){ int v = G[x][i]; if(!dfn[v]){ tarjan(v,father); low[x] = min(low[x],low[v]); if(dfn[x] <= low[v] && x != father) cut[x] = 1; if(x == father) child++; } low[x] = min(low[x],dfn[v]); } if(x == father && child >= 2) cut[x] = 1; } int main(){ int x,y; scanf("%d%d",&n,&m); for(int i = 1;i <= m;i++){ scanf("%d%d",&x,&y); G[x].push_back(y); G[y].push_back(x); } int ans = 0; for(int i = 1;i <= n;i++) if(!dfn[i]) tarjan(i,i); for(int i = 1;i <= n;i++){ if(cut[i]) ans++; } printf("%d ",ans); for(int i = 1;i <= n;i++){ if(cut[i]) printf("%d ",i); } return 0; }