kuangbin最小生成树

POJ - 1251 Jungle Roads

思路

Prim最小生成树模板题

代码

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

using namespace std;
typedef long long LL;
const int N = 200, M = N * N;
int g[N][N];
int dist[N];
int n;
bool st[N];


LL prim() {
    LL res = 0;

    memset(dist, 0x3f, sizeof dist);
    for (int i = 0; i < n; i ++) {
        int t = -1;
        for (int j = 0; j < n; j ++) {
            if(!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;
        }

        if(i)
            res += dist[t];

        st[t] = true;

        for (int j = 0; j < n; j ++) {
            dist[j] = min(dist[j], g[t][j]);
        }
    }

    return res;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    while(cin >> n && n) {
        memset(g, 0x3f, sizeof g);
        memset(st, false, sizeof st);

        for (int i = 0; i < n - 1; i ++) {
            char A[2];
            cin >> A;
            int a = A[0] - 'A';
            int m;
            cin >> m;
            while(m --) {
                char B[2];
                int c;
                cin >> B >> c;
                int b = B[0] - 'A';
                g[a][b] = g[b][a] = min(g[a][b], c);
            }
        }

         LL ans = prim();
        cout << ans << endl;
    }

    return 0;
}

POJ - 1287 Networking

思路

Prim模板

代码

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

using namespace std;
typedef long long LL;

const int N = 55;
int g[N][N];
int dist[N];
bool st[N];
int n, m;

LL prim() {
    LL res = 0;

    memset(dist, 0x3f, sizeof dist);
    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;
        }

        if(i)
            res += dist[t];

        st[t] = true;

        for (int j = 1; j <= n; j ++) {
            dist[j] = min(dist[j], g[t][j]);
        }
    }

    return res;
}

int main() {
    while(cin >> n >> m && n) {
        memset(g, 0x3f, sizeof g);
        memset(st, false, sizeof st);
        
        while(m --) {
            int a, b, c;
            cin >> a >> b >> c;
            g[a][b] = g[b][a] = min(g[a][b], c);
        }
        LL ans = prim();
        cout << ans << endl;
    }

    return 0;
}

POJ - 2421 Constructing Roads

题意

第一部分输入边的信息
第二部分输入 一开始哪些边已经建好了

思路

直接套Kruscal好写点

心得

函数的重构学到了
(再也不怕不支持c++11了，c++11直接写大括号赋值)

代码

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

using namespace std;

const int N = 110, M = N * N;
const int inf = 0x3f3f3f;
typedef long long LL;

struct Edge {
    int a, b, w;
    Edge(int aa, int bb, int ww) {
        a = aa, b = bb, w = ww;
    }
    Edge() {
        a = 0, b = 0, w = inf;
    }

    bool operator < (const Edge &W)const {
        return w < W.w;
    }
}edges[M];
int p[N];
int cnt;

int find(int x) {
    if(p[x] != x)
        p[x] = find(p[x]);
    return p[x];
}

void init()
{
    for (int i = 1; i < N; i ++)
        p[i] = i;
    cnt = 0;
}

LL Kruscal() {
    LL res = 0;

    sort(edges, edges + cnt);

    for (int i = 0; i < cnt; i ++) {
        int a = edges[i].a, b = edges[i].b, w = edges[i].w;
        int fa = find(a), fb = find(b);
        if(fa != fb) {
            res += w;
            p[fa] = fb;
        }
    }

    return res;
}

int main() {
    int n;
    cin >> n;
    init();
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= n; j ++) {
            int w;
            cin >> w;
            edges[cnt++] = Edge{i, j, w};
        }

    int m;
    cin >> m;
    while(m --) {
        int a, b;
        cin >> a >> b;
        int fa = find(a), fb = find(b);
        if(fa != fb)
            p[fa] = fb;
    }

    LL ans = Kruscal();

    cout << ans << endl;

    return 0;
}

POJ - 2349 Arctic Network

题意

给出k个卫星, 求最小的D使得村庄的连接都能连通
k个卫星的作用：可以删除最小生成树中k-1条边
D就是删边之后最小生成树中最大的边

思路

也挺简单，就是求Kruscal记录边的权值，输出w[m-k-1]即可

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 510, M = N * N;

int k, m;

PII point[M];
int cnt1;
int cnt2;
double weight[M];
int p[N];

struct Edge {
    int a, b;
    double w;
    Edge(int aa, int bb, double ww) {
        a = aa, b = bb, w = ww;
    }
    Edge() {
        a = 0, b = 0, w = 0;
    }
    bool operator < (const Edge& W)const {
        return w < W.w;
    }
} edges[M];

void init() {
    cnt1 = cnt2 = 0;
    for (int i = 0; i < N; i ++)
        p[i] = i;
}

int find(int x) {
    if(p[x] != x)
        return p[x] = find(p[x]);
    return p[x];
}
double get_dis(int x1, int y1, int x2, int y2) {
    return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

double Kruscal() {
    sort(edges, edges + cnt1);
    for (int i = 0; i < cnt1; i ++) {
        int a = edges[i].a, b = edges[i].b;
        double dis = edges[i].w;
        int fa = find(a), fb = find(b);
        if(fa != fb) {
            weight[cnt2++] = dis;
            p[fa] = fb;
        }
    }

    return weight[m - k - 1];
}

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {
        init();
        scanf("%d%d", &k, &m);
        for (int i = 0; i < m; i ++)
        {
            int x, y;
            scanf("%d%d", &point[i].x, &point[i].y);
        }

        for (int i = 0; i < m; i ++)
            for (int j = i + 1; j < m; j ++) {
                double dis = get_dis(point[i].x, point[i].y, point[j].x, point[j].y);
                edges[cnt1++] = Edge{i, j, dis};
            }

        double ans = Kruscal();
        printf("%.2f\n", ans);
    }

    return 0;
}