Acwing-----演算法基礎課之第三章搜尋與圖論（二）

阿新 • • 發佈：2020-09-20

最短路：

1.單源最短路
- 所有邊權都是正數：樸素Dijkstra演算法（O(\(n^2\))）、堆優化Dijkstra演算法（O(\(mlongn\))）
- 存在負權邊：Bellman-Ford演算法（O(\(nm\))）、SPFA演算法（一般O(\(m\))，最壞O(\(nm\))）
2.多源匯最短路：Floyd演算法（O(\(n^3\))）

849. Dijkstra求最短路 I

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

// 樸素版Dijkstra
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = 510;

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

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[t] > dist[j])) 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 a, b, c;
        cin >> a >> b >> c;
        g[a][b] = min(g[a][b], c);
    }
    cout << dijkstra() << endl;
    return 0;
}

850. Dijkstra求最短路 II

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

// 堆優化
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>

using namespace std;
typedef pair<int, int> PII;
const int N = 1e6 + 10;
int n, m, h[N], e[N], ne[N], dist[N], w[N], idx;
bool st[N];

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

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

853. 有邊數限制的最短路

連結：https://www.acwing.com/problem/content/855/

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

using namespace std;
const int N = 510, M = 100010;
int n, m, k, dist[N], backup[N];

struct edge {
    int a, b, c;
} edges[M];

void bellman_ford() {
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    
    for (int i = 0; i < k; ++i) {
        memcpy(backup, dist, sizeof dist);
        for (int j = 0; j < m; ++j) {
            auto e = edges[j];
            dist[e.b] = min(dist[e.b], backup[e.a] + e.c);
        }
    }
}

int main() {
    cin >> n >> m >> k;
    for (int i = 0 ;i < m; ++i) {
        int a, b, c;
        cin >> a >> b >> c;
        edges[i] = {a, b, c};
    }
    bellman_ford();
    
    if (dist[n] > 0x3f3f3f3f / 2) puts("impossible");
    else cout << dist[n] << endl;
    return 0;
}

851. spfa求最短路

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

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

using namespace std;
const int N = 100010;
int n, m, e[N], ne[N], h[N], w[N], idx, dist[N];
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 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;
                }
            }
        }
    }
}

int main() {
    cin >> n >> m;
    memset(h, -1, sizeof h);
    
    while (m--) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }
    
    spfa();
    
    if (dist[n] == 0x3f3f3f3f) puts("impossible");
    else cout << dist[n] << endl;
    return 0;
}

852. spfa判斷負環

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

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

using namespace std;
const int N = 2010, M = 10010;
int n, m, e[M], h[N], w[M], dist[N], ne[M], idx, cnt[N];
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 spfa() {
    queue<int> q;
    
    for (int i = 1; i <= n; ++i) {
        st[i] = true;
        q.push(i);
    }
    
    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] = cnt[t] + 1;
                
                if (cnt[j] >= n) return true;
                if (!st[j]) {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    return false;
}

int main() {
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while (m--) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }
    if (spfa()) puts("Yes");
    else puts("No");
    return 0;
}

854. Floyd求最短路

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

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 210, INF = 1e9;

int n, m, q;
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 >> q;
    
    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--) {
        int a, b, c;
        cin >> a >> b >> c;
        d[a][b] = min(d[a][b], c);
    }
    
    floyd();

    while (q--) {
        int a, b;
        cin >> a >> b;
        if (d[a][b] > INF / 2) puts("impossible");
        else cout << d[a][b] << endl;
    }
    
    return 0;
}