acwing算法提高之图论--单源最短路的综合应用

YMWM_2024-04-02 17:14

目录

  • [1 介绍](#1 介绍)
  • [2 训练](#2 训练)

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、二分、拓扑排序、动态规划等等）结合起来解题。

2 训练

题目1 ：1135新年好

C++代码如下，

cpp 复制代码 
//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;   
}
cpp 复制代码 
//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，

cpp 复制代码 
#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 ：342道路与航线

C++代码如下，

cpp 复制代码 
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 25010, M = 150010, INF = 0x3f3f3f3f;
int n, mr, mp, S;
int id[N];
int h[N], e[M], w[M], ne[M], idx;
int dist[N], din[N];
vector<int> block[N];
int bcnt;
bool st[N];
queue<int> q;

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

void dfs(int u, int bid) {
    id[u] = bid, block[bid].push_back(u);
    
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (!id[j]) {
            dfs(j, bid);
        }
    }
}

void dijkstra(int bid) {
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    
    for (auto u : block[bid]) {
        heap.push({dist[u], u});
    }
    
    while (heap.size()) {
        auto t = heap.top();
        heap.pop();
        
        int ver = t.y, distance = t.x;
        if (st[ver]) continue;
        st[ver] = true;
        
        for (int i = h[ver]; ~i; i = ne[i]) {
            int j = e[i];
            if (id[j] != id[ver] && --din[id[j]] == 0) q.push(id[j]);
            if (dist[j] > dist[ver] + w[i]) {
                dist[j] = dist[ver] + w[i];
                if (id[j] == id[ver]) heap.push({dist[j], j});
            }
        }
    }
}

void topsort() {
    memset(dist, 0x3f, sizeof dist);
    dist[S] = 0;
    
    for (int i = 1; i <= bcnt; ++i) {
        if (!din[i]) {
            q.push(i);
        }
    }
    
    while (q.size()) {
        int t = q.front();
        q.pop();
        dijkstra(t);
    }
    
}

int main() {
    cin >> n >> mr >> mp >> S;
    memset(h, -1, sizeof h);
    
    while (mr--) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c), add(b, a, c);
    }
    
    for (int i = 1; i <= n; ++i) {
        if (!id[i]) {
            bcnt++;
            dfs(i, bcnt);
        }
    }
    
    while (mp--) {
        int a, b, c;
        cin >> a >> b >> c;
        din[id[b]]++;
        add(a, b, c);
    }
    
    topsort();
    
    for (int i = 1; i <= n; ++i) {
        if (dist[i] > INF / 2) cout << "NO PATH" << endl;
        else cout << dist[i] << endl;
    }
    
    return 0;
}

题目4 ：341最优贸易

C++代码如下，

cpp 复制代码 
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 100010, M = 2000010;
int n, m;
int w[N];
int hs[N], ht[N], e[M], ne[M], idx;
int dmin[N], dmax[N];
int q[N];
bool st[N];

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

void spfa(int h[], int dist[], int type) {
    int hh = 0, tt = 1;
    if (type == 0) {
        memset(dist, 0x3f, sizeof dmin);
        dist[1] = w[1];
        q[0] = 1;
    } else {
        memset(dist, -0x3f, sizeof dmax);
        dist[n] = w[n];
        q[0] = n;
    }
    
    while (hh != tt) {
        int t = q[hh++];
        if (hh == N) hh = 0;
        
        st[t] = false;
        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i];
            if (type == 0 && dist[j] > min(dist[t], w[j]) || type == 1 && dist[j] < max(dist[t], w[j])) {
                if (type == 0) dist[j] = min(dist[t], w[j]);
                else dist[j] = max(dist[t], w[j]);
                
                if (!st[j]) {
                    q[tt++] = j;
                    if (tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", &w[i]);
    
    memset(hs, -1, sizeof hs);
    memset(ht, -1, sizeof ht);
    
    while (m--) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(hs, a, b), add(ht, b, a);
        if (c == 2) add(hs, b, a), add(ht, a, b);
    }
    
    spfa(hs, dmin, 0);
    spfa(ht, dmax, 1);
    
    int res = 0;
    for (int i = 1; i <= n; ++i) res = max(res, dmax[i] - dmin[i]);
    
    printf("%d\n", res);
    
    return 0;
}
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