圖論算法歸納（Dijkstra+SPFA+Floyd+Prim+Kruskal+二分圖）

一、Dijkstra鄰接矩陣求最短路

題目連結：https://www.acwing.com/problem/content/851/

代碼：

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

const int N = 510;
int g[N][N];
int dist[N];
bool st[N];
int n,m;

int Dijkstra(){
    memset(dist,0x3f,sizeof dist);
    dist[1] = 0;
    for(int i = 0;i<n;i++){
        int t = -1;
        for(int j = 1;j<=n;j++){
            if(!st[j]&&(t==-1||dist[j]<dist[t]))t = j;
        }
        st[t] = true;
        for(int j = 1;j<=n;j++)
            dist[j] = min(dist[j],dist[t]+g[t][j]);
    }
    if(dist[n]==0x3f3f3f3f)return -1;
        return dist[n];
}


int main(){
    cin>>n>>m;
    memset(g,0x3f,sizeof g);
    while(m--){
        int x,y,z;
        cin>>x>>y>>z;
        g[x][y] = min(g[x][y],z);
    }
    cout<<Dijkstra()<<endl;
    return 0;
}
           

二、Dijkstra鄰接表求最短路

題目連結：https://www.acwing.com/problem/content/852/

代碼：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>

using namespace std;
typedef pair<int, int> PII; //整數對

const int N = 150010;
int h[N], e[N], ne[N], idx, w[N];
int dist[N];
bool st[N];
int n, m;

void add(int f,int u,int c){
    ne[idx] = h[f];
    w[idx] = c;
    e[idx] = u;
    h[f] = idx++;
}

int dijkstra(){
    memset(dist,0x3f,sizeof dist);
    memset(st,false,sizeof st);
    dist[1] = 0;
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0,1});
    while(heap.size()){
        PII u = heap.top();
        heap.pop();
        int distance = u.first,v = u.second;
        if(st[v])continue;
        st[v] = true;
        
        for(int i = h[v];i!=-1;i = ne[i]){
            int j = e[i];
            if(dist[j]>distance+w[i]){
                dist[j] = distance +w[i];
                heap.push({dist[j],j});
            }
        }
    }
    
    if(dist[n]==0x3f3f3f3f)return -1;
        return dist[n];
}


int main() {
    memset(h, -1, sizeof h);
    cin >> n >> m;
    while (m--) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }
    cout << dijkstra() << endl;
    return 0;
}
           

三、SPFA算法求最短路

題目連結：https://www.acwing.com/problem/content/853/

代碼：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>

using namespace std;
const int N=100010;
int h[N],e[N],ne[N],idx,w[N];
int dist[N];
bool st[N];
 int n,m;
void add(int a,int b,int c){
    e[idx]=b;
  w[idx]=c;
  ne[idx]=h[a];
  h[a]=idx++;
}


int spfa(){
    memset(dist,0x3f,sizeof dist);
     dist[1]=0;
     queue<int>q;
     q.push(1);
     st[1]=true;
    while(q.size()){
        int t = q.front();
        q.pop();
        st[t] = false;
        for(int i = h[t];i!=-1;i = ne[i]){
            int j = e[i];
            if(dist[j]>dist[t]+w[i]){
                dist[j] = dist[t]+w[i];
                if(!st[j]){
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    return dist[n];
}

int main(){
    cin>>n>>m;
    memset(h,-1,sizeof h);
    while(m--){
        int a,b,c;
        cin>>a>>b>>c;
        add(a,b,c);
    }

    int t=spfa();
    if(t==0x3f3f3f3f)puts("impossible");
    else cout<<t<<endl;
    return 0;
}

           

四、SPFA判負環

題目連結：https://www.acwing.com/problem/content/854/

代碼：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>

using namespace std;
const int N=10010;
int h[N],e[N],ne[N],idx,w[N];
int dist[N];
bool st[N];
 int n,m;
 int cnt[N];
void add(int a,int b,int c){
    e[idx]=b;
  w[idx]=c;
  ne[idx]=h[a];
  h[a]=idx++;
}


int spfa(){
    memset(dist,0x3f,sizeof dist);
    dist[1]=0;
    
    queue<int>q;
    for(int i = 1;i<=n;i++)q.push(i),st[i] = true;
    st[1]=true;
    while(q.size()){
        int t = q.front();
        q.pop();
        st[t] = false;
        for(int i = h[t];i!=-1;i = ne[i]){
            int j = e[i];
            if(dist[j]>dist[t]+w[i]){
                dist[j] = dist[t]+w[i];
                cnt[j] ++;
                if(cnt[j]>=n)return 1;
                if(!st[j]){
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    return 0;
}

int main(){
    cin>>n>>m;
    memset(h,-1,sizeof h);
    while(m--){
        int a,b,c;
        cin>>a>>b>>c;
        add(a,b,c);

    }
    int t=spfa();
    if(t==1)cout<<"Yes";
    else cout<<"No";
    return 0;
}
           

五、Floyd算法求多源最短路

題目連結：https://www.acwing.com/problem/content/856/

代碼：

#include <iostream>
using namespace std;

const int N = 210, M = 2e+10, INF = 1e9;

int n, m, k, x, y, z;
int d[N][N];

void floyd() {
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                d[i][j] = min(d[i][j], d[i][k] + d[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) d[i][j] = 0;
            else d[i][j] = INF;
    while(m--) {
        cin >> x >> y >> z;
        d[x][y] = min(d[x][y], z);
        //注意儲存最小的邊
    }
    floyd();
    while(k--) {
        cin >> x >> y;
        if(d[x][y] > INF/2) puts("impossible");
        //由于有負權邊存在是以約大過INF/2也很合理
        else cout << d[x][y] << endl;
    }
    return 0;
}
           

六、Prim算法求最小生成樹

題目連結：https://www.acwing.com/problem/content/1142/

代碼：

#include <bits/stdc++.h>
using namespace std;

const int N = 105;
int n;
int w[N][N],dist[N];
bool st[N];


int prim(){
    int res = 0;
    memset(dist,0x3f,sizeof dist);
    dist[1] = 0;
    for(int i = 0;i<n;i++){
        int t = -1;
        for(int j = 1;j<=n;j++){
            if(!st[j]&&(t==-1||dist[t]>dist[j]))t = j;
        }
        res += dist[t];
        st[t] = true;
        for(int j = 1;j<=n;j++)dist[j] = min(dist[j],w[t][j]);
    }
    return res;
}

int main(){
    cin>>n;
    for(int i = 1;i<=n;i++)
        for(int j = 1;j<=n;j++)
            cin>>w[i][j];
    cout<<prim()<<endl;
    
    return 0;
}
           

七、Kruskal算法求最小生成樹

題目連結：https://www.acwing.com/problem/content/861/

代碼：

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5+10, M = 2e5+10, INF = 0x3f3f3f3f;

int n, m;
int p[N];//祖宗節點

struct Edge {
    int u, v, w;

    bool operator < (const Edge & T) const {
        return w < T.w;
    }

}edges[M];

int find(int x) {
    //并查集核心
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int kruskal() {
    sort(edges, edges+m);
    for(int i = 1; i <= n; i++) p[i] = i; //初始化并查集

    int res = 0, cnt = 0;
    //res 最小生成樹中的權重之和
    //cnt 目前加了多少條邊

    for(int i = 0; i < m; i++) {
        auto t = edges[i];
        t.u = find(t.u), t.v = find(t.v);
        if(t.u != t.v) {
            p[t.u] = t.v;
            res += t.w;
            cnt++;
        }
    }

    if(cnt < n - 1) return INF;
    return res;
}

int main() {
    cin >> n >> m;
    int u, v, w;

    for(int i = 0; i < m; i++) {
        cin >> u >> v >> w;
        edges[i] = {u, v, w};
    }

    int x = kruskal();
    if(x > INF/2) puts("impossible");
    else cout << x << endl;
}
           

七、染色法求二分圖

題目連結：https://www.acwing.com/problem/content/862/

代碼：

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5+10,M = 2*N;
int n,m;
int h[N],e[M],ne[M],idx;
int st[N];
void add(int f ,int u){
    ne[idx] = h[f];
    e[idx] = u;
    h[f] = idx++;
}



bool bfs(int t){
    queue<pair<int,int>>q;
    q.push({t,1});
    st[t] = 1;
    while(q.size()){
        auto u = q.front();
        q.pop();
        int color = u.second;
        int r = u.first;
        for(int i = h[r];i!=-1;i = ne[i]){
            int j = e[i];
            if(st[j]){
                if(st[j]==color)return false;
            }else{
                q.push({j,3-color});
                st[j] = 3-color;
            }
        }
    }
    return true;
}

int main(){
    cin>>n>>m;
    memset(h,-1,sizeof h);
    while(m--){
        int a,b;
        cin>>a>>b;
        add(a,b);
        add(b,a);
    }
    
    bool flag = true;
    
    for(int i = 1;i<=n;i++){
        if(!st[i]){
            if(!bfs(i)){
                flag = false;
                break;
            }
        }
    }
    
    if(flag)cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
    return 0;
}

           

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