HDU 1102 Constructing Roads (最小生成树)

最小生成树模板（嗯……在kuangbin模板里面抄的……）

最小生成树(prim)

/** Prim求MST

  * 耗费矩阵cost[][]，标号从0开始，0~n-1

  * 返回最小生成树的权值，返回-1表示原图不连通

  */

const int INF = 0x3f3f3f3f;

const int MAXN = 110;



bool vis[MAXN];

int lowc[MAXN];

int map[MAXN][MAXN];



int Prim(int cost[][MAXN], int n)

{

    int ans = 0;

    memset(vis, false, sizeof(vis));

    vis[0] = true;

    for (int i = 1; i < n; ++i) lowc[i] = cost[0][i];

    for (int i = 1; i < n; ++i) {

        int minc = INF;

        int p = 1;

        for (int j = 0; j < n; ++j)

            if (!vis[j] && minc > lowc[j]) {

                minc = lowc[j];

                p = j;

            }

        if (minc == INF) return -1;

        ans += minc;

        vis[p] = true;

        for (int j = 0; j < n; ++j)

            if (!vis[j] && lowc[j] > cost[p][j])

                lowc[j] = cost[p][j];

    }

    return ans;

}

感觉Prim和Dijkstra有点像。写起来挺简单的。

我一开始的想法是把每个q中的a，b设为已访问节点，后来发现不对，例如ab连，dc连，但它们并不是全部相连的。

好吧其实此题就是把已建好两点之间距离设为0.

//Problem : 1102 ( Constructing Roads )     Judge Status : Accepted

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;



const int N = 105;

const int INF = 3000;



int map[N][N];

int vis[N];

int dis[N];



int Prim(int n)

{

    int ans = 0;

    memset(vis, 0, sizeof(vis));

    for (int j = 1; j <= n; ++j) {

        dis[j] = map[1][j];

    }

    for (int k = 1; k <= n; ++k) {

        int minc = INF;

        int p = 1;

        for (int i = 1; i <= n; ++i) {

            if (!vis[i] && dis[i] < minc) {

                minc = dis[i];

                p = i;

            }

        }

        if (minc == INF) return -1;

        ans += minc;

        vis[p] = 1;

        for (int j = 1; j <= n; ++j) {

            if (map[p][j] < dis[j] && !vis[j])

                dis[j] = map[p][j];

        }

    }

    return ans;

}



int main()

{

    int n, q;

    while (cin >> n) {

        for (int i = 1; i <= n; ++i)

            for (int j = 1; j <= n; ++j)

                cin >> map[i][j];

        cin >> q;

        int a, b;

        for (int i = 0; i < q; ++i) {

            cin >> a >> b;

            map[a][b] = map[b][a] = 0;

        }

        cout << Prim(n) << endl;

    }

    return 0;

}