acwing算法提高之图论--单源最短路的扩展应用

1 介绍

本专题用来记录使用。。。。

2 训练

题目1：1137选择最佳线路

C++代码如下，

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

using namespace std;

const int N = 1010, M = 20010;
int n, m;
int dist[N];
int st[N];
vector<vector<pair<int,int>>> g;
vector<int> snodes;
int enode; 
int w;

void spfa() {
    //cout << " ====== " << endl;
    
    memset(dist, 0x3f, sizeof dist);
    memset(st, 0, sizeof st);
    
    queue<int> q;
    for (auto snode : snodes) {
        //cout << "snode = " << snode << endl;
        dist[snode] = 0;
        q.push(snode);
        st[snode] = true; //已经在队列中了
    }
    
    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 main() {
    while (cin >> n >> m >> enode) {
        g.clear();
        g.resize(n + 10);
        for (int i = 0; i < m; ++i) {
            int a, b, c;
            cin >> a >> b >> c;
            g[a].emplace_back(b, c);
        }
        
        cin >> w;
        snodes.resize(w);
        for (int i = 0; i < w; ++i) cin >> snodes[i];
        
        spfa();
        
        if (dist[enode] == 0x3f3f3f3f) cout << "-1" << endl;
        else cout << dist[enode] << endl;
    }
    
    return 0;
}

题目2：1131拯救大兵瑞恩

C++代码如下，

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

#define x first
#define y second

using namespace std;

typedef pair<int,int> PII;

const int N = 11, M = 360, P = 1 << 10;

int n, m, k, p;
int h[N * N], e[M], w[M], ne[M], idx;
int g[N][N], key[N * N];
int dist[N * N][P];
bool st[N * N][P];

set<PII> edges;

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

void build() {
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            for (int u = 0; u < 4; ++u) {
                int x = i + dx[u], y = j + dy[u];
                if (!x || x > n || !y || y > m) continue;
                int a = g[i][j], b = g[x][y];
                if (!edges.count({a,b})) add(a, b, 0);
            }
        }
    }
    return;
}

int bfs() {
    memset(dist, 0x3f, sizeof dist);
    dist[1][0] = 0;
    
    deque<PII> q;
    q.push_back({1, 0});
    
    while (q.size()) {
        PII t = q.front();
        q.pop_front();
        
        if (st[t.x][t.y]) continue;
        st[t.x][t.y] = true;
        
        if (t.x == n * m) return dist[t.x][t.y];
        
        if (key[t.x]) {
            int state = t.y | key[t.x];
            if (dist[t.x][state] > dist[t.x][t.y]) {
                dist[t.x][state] = dist[t.x][t.y];
                q.push_front({t.x, state});
            }
        }
        
        for (int i = h[t.x]; ~i; i = ne[i]) {
            int j = e[i];
            if (w[i] && !(t.y >> w[i] - 1 & 1)) continue;
            if (dist[j][t.y] > dist[t.x][t.y] + 1) {
                dist[j][t.y] = dist[t.x][t.y] + 1;
                q.push_back({j, t.y});
            }
        }
    }
    return -1;
}

int main() {
    cin >> n >> m >> p >> k;
    
    for (int i = 1, t = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            g[i][j] = t++;
        }
    }
    
    memset(h, -1, sizeof h);
    while (k--) {
        int x1, y1, x2, y2, c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        int a = g[x1][y1], b = g[x2][y2];
        
        edges.insert({a, b}), edges.insert({b, a});
        if (c) add(a, b, c), add(b, a, c);
    }
    
    build();
    
    int s;
    cin >> s;
    while (s--) {
        int x, y, c;
        cin >> x >> y >> c;
        key[g[x][y]] |= 1 << c - 1;
    }
    
    cout << bfs() << endl;
    
    return 0;
}

题目3：1134最短路计数

C++代码如下，

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

using namespace std;

const int N = 1e5 + 10, mod = 100003;
int n, m;
vector<vector<int>> g;
int dist[N];
int cnt[N];

void bfs() {
    memset(dist, 0x3f, sizeof dist);
    
    queue<int> q;
    q.push(1);
    cnt[1] = 1;
    dist[1] = 0;
    
    while (!q.empty()) {
        auto t = q.front();
        q.pop();
        
        for (auto b : g[t]) {
            if (dist[b] > dist[t] + 1) {
                dist[b] = dist[t] + 1;
                cnt[b] = cnt[t];
                q.push(b);
            } else if (dist[b] == dist[t] + 1) {
                cnt[b] = (cnt[b] + cnt[t]) % mod;
            }
        }
    }
    
    return;
}

int main() {
    cin >> n >> m;
    g.resize(n + 10);
    for (int i = 0; i < m; ++i) {
        int a, b;
        cin >> a >> b;
        g[a].emplace_back(b);
        g[b].emplace_back(a);
    }
    
    bfs();
    
    for (int i = 1; i <= n; ++i) {
        cout << cnt[i] << endl;
    }
    
    return 0;
}

题目4：383观光

C++代码如下，

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

using namespace std;

typedef tuple<int,int,int> TIII; //distance,node,type

const int N = 1010;
int dist[N][2];
int cnt[N][2];
bool st[N][2];
int n, m;
int S, F;
vector<vector<pair<int,int>>> g;

void dijkstra() {
    memset(dist, 0x3f, sizeof dist);
    memset(cnt, 0, sizeof cnt);
    memset(st, 0, sizeof st);
    
    priority_queue<TIII, vector<TIII>, greater<TIII>> hp;
    
    dist[S][0] = 0, cnt[S][0] = 1;
    hp.push({0, S, 0});
    
    while (!hp.empty()) {
        int distance, node, type;
        tie(distance, node, type) = hp.top();
        hp.pop();
        
        int count = cnt[node][type];
        
        if (st[node][type]) continue;
        
        st[node][type] = true;
        
        for (auto [b, w] : g[node]) {
            if (dist[b][0] > distance + w) {
                dist[b][1] = dist[b][0], cnt[b][1] = cnt[b][0];
                hp.push(make_tuple(dist[b][1], b, 1));
                
                dist[b][0] = distance + w, cnt[b][0] = count;
                hp.push(make_tuple(dist[b][0], b, 0));
            } else if (dist[b][0] == distance + w) cnt[b][0] += count;
            else if