acwing算法提高之图论-爱代码爱编程

1 介绍

本专题用来介绍使用最短路算法（spfa或dijkstra）与其它算法（dfs等等）结合起来解题。

2 训练

题目1：1135新年好

C++代码如下，

//版本1，使用vector来存储图，会有超时风险
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

const int N = 50010;
int n, m;
int dist[6][N];
bool st[N];
vector<vector<pair<int, int>>> g; //使用vector可能有超时分险
int source[6];

void spfa(int start, int dist[]) {
    memset(dist, 0x3f, N * 4);
    memset(st, 0, sizeof st);
    
    dist[start] = 0;
    queue<int> q;
    q.push(start);
    st[start] = true;
    
    while (!q.empty()) {
        auto t = q.front();
        q.pop();
        st[t] = false;
        
        for (auto [b, w] : g[t]) {
            if (dist[b] > dist[t] + w) {
                dist[b] = dist[t] + w;
                if (!st[b]) {
                    q.push(b);
                    st[b] = true;
                }
            }
        }
        
    }
    
    return;
}

int dfs(int u, int start, int distance) {
    if (u > 5) return distance;
    
    int res = 0x3f3f3f3f;
    for (int i = 1; i <= 5; ++i) {
        if (!st[i]) {
            int b = source[i];
            st[i] = true;
            res = min(res, dfs(u + 1, i, distance + dist[start][b]));
            st[i] = false; 
        }
    }
    return res;
}

int main() {
    cin >> n >> m;
    g.resize(n + 10);
    
    source[0] = 1;
    for (int i = 1; i <= 5; ++i) cin >> source[i];
    
    for (int i = 0; i < m; ++i) {
        int a, b, c;
        cin >> a >> b >> c;
        g[a].emplace_back(b, c);
        g[b].emplace_back(a, c);
    }
    
    for (int i = 0; i < 6; ++i) spfa(source[i], dist[i]);
    
    memset(st, 0, sizeof st);
    int res = dfs(1, 0, 0);
    
    cout << res << endl;
    
    return 0;   
}

//版本2，使用h,e,ne,w数组来存储图
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 50010, M = 200010, INF = 0x3f3f3f3f;

int n, m;
int h[N], e[M], w[M], ne[M], idx;
int q[N], dist[6][N];
int source[6];
bool st[N];

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

void dijkstra(int start, int dist[]) {
    memset(dist, 0x3f, N * 4);
    dist[start] = 0;
    memset(st, 0, sizeof st);
    
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, start});
    
    while (heap.size()) {
        auto t = heap.top();
        heap.pop();
        
        int ver = t.second;
        if (st[ver]) continue;
        st[ver] = true;
        
        for (int i = h[ver]; ~i; i = ne[i]) {
            int j = e[i];
            if (dist[j] > dist[ver] + w[i]) {
                dist[j] = dist[ver] + w[i];
                heap.push({dist[j], j});
            }
        }
    }
}

int dfs(int u, int start, int distance) {
    if (u > 5) return distance;
    
    int res = INF;
    for (int i = 1; i <= 5; ++i) {
        if (!st[i]) {
            int next = source[i];
            st[i] = true;
            res = min(res, dfs(u + 1, i, distance + dist[start][next]));
            st[i] = false; 
        }
    }
    return res;
}

int main() {
    scanf("%d%d", &n, &m);
    source[0] = 1;
    for (int i = 1; i <= 5; ++i) scanf("%d", &source[i]);
    
    memset(h, -1, sizeof h);
    while (m--) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c), add(b, a, c);
    }
    
    for (int i = 0; i < 6; ++i) dijkstra(source[i], dist[i]);
    
    memset(st, 0, sizeof st);
    printf("%d\n", dfs(1, 0, 0));
    
    return 0;
}

题目2：340通信线路

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <deque>

using namespace std;

const int N = 1010, M = 20010;
int n, m, k;
int h[N], e[M], w[M], ne[M], idx;
int dist[N];
deque<int> q;
bool st[N];

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

bool check(int bound) {
    memset(dist, 0x3f, sizeof dist);
    memset(st, 0, sizeof st);
    
    q.push_back(1);
    dist[1] = 0;
    
    while (q.size()) {
        int t = q.front();
        q.pop_front();
        
        if (st[t]) continue;
        st[t] = true;
        
        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i], x = w[i] > bound;
            if (dist[j] > dist[t] + x) {
                dist[j] = dist[t] + x;
                if (!x) q.push_front(j);
                else q.push_back(j);
            }
        }
    }
    return dist[n] <= k;
}

int main() {
    cin >> n >> m >> k;
    memset(h, -1, sizeof h);
    while (m--) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c), add(b, a, c);
    }
    
    int l = 0, r = 1e6 + 1;
    while (l < r) {
        int mid = l + r >> 1;
        if (check(mid)) r = mid;
        else l = mid + 1;
    }
    
    if (r == 1e6 + 1) cout << -1 << endl;
    else cout << r << endl;
    
    return 0;
}

题目3：