1. 程式人生 > >複習1-圖論模板

複習1-圖論模板

1.最短路

圖全為正權使用Dijkstra,有負權用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); } } } } }
SPFA
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]}); } } } }
Dijkstra+堆優化
/*
    Bellman-Ford演算法 題解上的
*/

#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){ //迴圈大法好
        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;
}
Bellman-Ford

 

2.最小生成樹

主要掌握Kruskal演算法,Prim演算法稍加了解即可

/*
     適用於稀疏圖 
*/
#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\n",Kruskal());
    return 0;
}
Kruskal演算法
/*
    適用於稠密圖(然而沒啥用)
*/

#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\n",Prim());
    return 0;
}
Prim演算法

 

3.LCA

主要理解倍增法,樹剖也可以瞭解,Tarjan就算了

/*
    倍增寫法 常數略大
    我覺得不太好理解 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\n",lca(a,b));
    }
    return 0;
}
倍增
/*
    這種樹剖寫法好理解(好背)
    雖然程式碼比倍增略長 但是也比倍增快
    簡直完美2333333
    所以以後就用這種方法辣
*/

#include <bits/stdc++.h>

using namespace std;

const int maxn = 500005;

int fa[maxn],top[maxn],id[maxn],son[maxn],depth[maxn],size[maxn];//樹剖要用的所有陣列
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\n",lca(a,b));
    }
    return 0;
}
樹剖
/*
    Tarjan演算法(題解) 常數挺大的,不推薦,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後再合併
    }
    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過
    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\n",ans[i]);
    }
    return 0;
}
Tarjan

 

4.二分圖最大匹配

要會

#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\n",ans);
    return 0;
}
匈牙利演算法

 

5.Tarjan

全部要會

vector<int> G[maxn];  //vector鄰接表存圖 
int pre[maxn],low[maxn],sccno[maxn],cnt,scccnt;
//pre[i]表示節點i被搜到的次序,lowlink[i]表示i及其後代能追溯到的最早的點v的
//pre[v]值,sccno[i]就是i所在的強連通分量的編號,dfs_clock表示第幾次dfs,
//scc_cnt表示找到的強連通分量序號的臨時值 
stack<int> S;//儲存dfs到的每一個點 

void dfs(int u)
{
    pre[u] = low[u] = ++cnt;//先把pre和lowlink初始化為dfs的時間戳 
    S.push(u);
    for(int i = 0;i < G[u].size();i++)//遍歷與點u相連的所有點 
    {
        int v = G[u][i];//取點 
        if(!pre[v]){//如果點v沒有遍歷過 
            dfs(v);//深搜 
            low[u] = min(low[u],low[v]);//向上合併lowlink的值 
        } else if(!sccno[v])//此時v已經搜過,但是不屬於任何一個scc,那麼就說明已經形成了環 
        {
            low[u] = min(low[u],pre[v]);
        }
    }
    if(low[u] == pre[u]){//如果u為最先搜到的點,它就是這個scc的根節點 
    scccnt++; 
    for(;;)
    {
        int x = S.top();S.pop();//取一個搜過的點 
        sccno[x] = scccnt;//它屬於這個強聯通分量 
        if(x == u) break;//直到棧中 
         }
    }
 } 

 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);
 }
求強連通分量
/*
     易得出狀態轉移為val[v] = max(val[v],val[pre] + a[v]) 
     於是通過tarjan演算法把所有環縮為一點,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為原來的圖,g2為縮點後的圖 

void add1(int from,int to){
    g1[++cnt].from = from;
    g1[cnt].to = to;
    g1[cnt].pre = head1[from];
    head1[from] = cnt;
} //對應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;
} //對應g2的add操作 

void tarjan(int x){
    low[x] = dfn[x] = ++scc_clock;//初始化為時間戳 
    stac[++top] = x;vis[x] = 1;//入棧、標誌陣列設為1 
    for(register int i = head1[x];i;i = g1[i].pre){//遍歷與x連線的點 
        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]);
        }
    }//一頓tarjan的操作 
    if(low[x] == dfn[x]){
        scc_amount++;
        int y;
        while(y = stac[top--]){//出棧 
            scc[y] = x;
            vis[y] = 0;//我也不知道為啥 
            if(x == y) break;
            a[x] += a[y];//加權值 
        }
    }
}

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演算法基本操作 
    }
    for(register int i = 1;i <= m;i++){
        x = scc[g1[i].from],y = scc[g1[i].to];
        if(x != y){//如果不屬於同一強連通分量,就連邊 
            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\n",ans);
    return 0;
}
縮點
/*
無向圖割點
對該圖進行一次 Tarjan 演算法(這裡注意在搜尋樹中把無向邊當做有向邊看。即LOW[u]=min(LOW[u],DFN[v])(v 是 u 的祖先)的條件變為(v 是 u 的祖先且 v 不是 u 的父親))這樣之後列舉搜尋樹上的所有邊(u,v),若存在 LOW[v]>=DNF[u],則 u 是割點。
無向圖割邊
對該圖進行一次 Tarjan 演算法(這裡注意在搜尋樹中把無向邊當做有向邊看。即LOW[u]=min(LOW[u],DFN[v])(v 是 u 的祖先)的條件變為(v 是 u 的祖先且 v 不是 u 的父親))這樣之後列舉搜尋樹上的所有邊(u,v),若存在 LOW[v]>DNF[u],則(u,v)為割邊。
*/
#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\n",ans);
    for(int i = 1;i <= n;i++){
        if(cut[i]) printf("%d ",i);
    }
    return 0;
}
求割點

 

6.樹上差分

7.拓撲排序