kruskal - dijskra - 堆優化的dijskra

寫在前面

整理了一份最小生成樹算法板子

題目 C - 掌握魔法の東東 I

東東在老家農村無聊，想種田。農田有$ n$ 塊，編號從 1~$n$。種田要灌氵
衆所周知東東是一個魔法師，他可以消耗一定的 MP 在一塊田上施展魔法，使得黃河之水天上來。他也可以消耗一定的 MP 在兩塊田的渠上建立傳送門，使得這塊田引用那塊有水的田的水。 $(1 \le n \le 3e2)$
黃河之水天上來的消耗是$W_i$，$i$ 是農田編號 $(1 \le W_i \le 1e5)$
建立傳送門的消耗是 $P_{ij}$，$i$、$j$是農田編號 $(1 \le P_{ij} \le 1e5, P_{ij} = P_{ji}, P_{ii} =0)$
東東爲所有的田灌氵的最小消耗

input

第1行：一個數$n$
第2行到第$n+1$行：數$w_i$
第$n+2$行到第$2n+1$行：矩陣即$p_{ij}$矩陣

output

東東最小消耗的MP值

sample input

4
5
4
4
3
0 2 2 2
2 0 3 3
2 3 0 4
2 3 4 0

Sample Output

9

kruskal

複雜度$O(mlogm)$

const int MAXN = 3e2 + 7;
int par[MAXN];
struct re {
    int x, y, w;
    bool operator<(const re& a) const { return w < a.w; }
} v[MAXN * MAXN];
int find(int x) { return par[x] == x ? x : par[x] = find(par[x]); }
int main() {
    int n;
    cin >> n;
    int cnt = 1;
    for (int i = 1; i <= n; i++) {
        v[cnt].x = 0;
        v[cnt].y = i;
        cin >> v[cnt].w;
        ++cnt;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            int w = get_num();
            if (i == j) continue;
            v[cnt].x = i;
            v[cnt].y = j;
            v[cnt].w = w;
            cnt++;
        }
    }
    sort(v + 1, v + cnt);
    int ans = 0, sum = 0;
    for (int i = 1; i <= n; i++) par[i] = i;
    for (int i = 1; i < cnt; i++) {
        if (sum == n) break;
        if (find(v[i].x) != find(v[i].y)) {
            ans += v[i].w;
            par[find(v[i].x)] = find(v[i].y);
            sum++;
        }
    }
    cout << ans;
    return 0;
}

dijskra

複雜度$O(n^2)$

const int MAXN = 3e2 + 7;
int v[MAXN][MAXN], vis[MAXN], d[MAXN];
int main() {
    int n;
    n = get_num();
    for (int i = 1; i <= n; i++) {
        v[0][i] = v[i][0] = get_num();
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            v[i][j] = get_num();
        }
    }
    for (int i = 0; i <= n; i++) {
        vis[i] = 0;
        d[i] = 1e9;
    }
    d[0] = 0;
    int ans = 0;
    for (int i = 0; i <= n; i++) {
        int sum = 1e9, k;
        for (int i = 0; i <= n; i++) {
            if (!vis[i] && d[i] < sum) {
                sum = d[i];
                k = i;
            }
        }
        ans += d[k];
        vis[k] = 1;
        for (int i = 0; i <= n; i++) {
            d[i] = min(d[i], v[k][i]);
        }
    }
    cout << ans;
    return 0;
}

堆優化的dijstra

複雜度$O((m+n)logn)$，若採用鄰接矩陣存圖時間複雜度爲$O(nlogn )$

const int MAXN = 3e2 + 7;
int v[MAXN][MAXN], vis[MAXN], d[MAXN];
struct re {
    int d, w;
    bool operator<(const re &a) const { return w > a.w; }
};
priority_queue<re> Q;
int main() {
    int n;
    n = get_num();
    for (int i = 1; i <= n; i++) {
        v[0][i] = v[i][0] = get_num();
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            v[i][j] = get_num();
        }
    }
    for (int i = 0; i <= n; i++) {
        vis[i] = 0;
        d[i] = 1e9;
    }
    d[0] = 0;
    Q.push({0, 0});
    int ans = 0;
    while (!Q.empty()) {
        re h = Q.top();
        Q.pop();
        if (vis[h.d]) continue;
        vis[h.d] = 1;
        ans += h.w;
        for (int i = 0; i <= n; i++) {
            if (d[i] > v[h.d][i]) {
                d[i] = v[h.d][i];
                Q.push({i, d[i]});
            }
        }
    }
    cout << ans;
    return 0;
}