【题解】POJ - 1251 Jungle Roads（最小生成树，Prim 算法，Kruskal 算法）

POJ - 1251 Jungle Roads

题目描述

The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems.

输入描述

The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above.

输出描述

The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit.

输入样例

9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18 G 55
D 1 E 44
E 2 F 60 G 38
F 0
G 1 H 35
H 1 I 35
3
A 2 B 10 C 40
B 1 C 20
0

输入样例

216
30

题目大意

给定一张图，需要建造铁路，求最小可以覆盖整张图的铁路花费。即最小生成树(Minimum Spanning Tree, MST)。

Prim 算法

Prim 算法进行选点来扩张生成树。

#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1e5 + 10;
const int inf = 0x3fffffff;
struct node {
     
    int to, len, next;
} edge[maxn << 1];
int head[maxn];
int n, k, w, cnt = 0;
char u, v;
int pre[maxn], lowcost[maxn];
bool vis[maxn];

void add_edge(int x, int y, int w) {
     
    cnt++;
    edge[cnt].len = w;
    edge[cnt].to = y;
    edge[cnt].next = head[x];
    head[x] = cnt;
}

int prim(int s) {
     
    fill(lowcost, lowcost + maxn, inf);
    memset(pre, -1, sizeof pre);
    memset(vis, false, sizeof vis);
    int ans = 0;

    pre[s] = lowcost[s] = 0;

    for (int i = 0; i < n; i++) {
     
        int mincost = inf, now = 0;
        for (int j = 0; j < 30; j++) {
     
            if (mincost > lowcost[j] && vis[j] == false) {
     
                mincost = lowcost[j];
                now = j;
            }
        }

        ans += lowcost[now];
        vis[now] = true;

        for (int i = head[now]; i; i = edge[i].next) {
     
            if (lowcost[edge[i].to] > edge[i].len) {
     
                lowcost[edge[i].to] = edge[i].len;
                pre[edge[i].to] = now;
            }
        }
    }
    return ans;
}

int main()
{
     
    while (cin >> n && n) {
     
        memset(head, 0, sizeof head);
        cnt = 0;
        for (int i = 0; i < n - 1; i++) {
     
            cin >> u >> k;
            for (int j = 0; j < k; j++) {
     
                cin >> v >> w;
                add_edge(u - 'A' + 1, v - 'A' + 1, w);
                add_edge(v - 'A' + 1, u - 'A' + 1, w);
            }
        }

        printf("%d\n", prim(u - 'A' + 1));
    }

    return 0;
}

Kruskal 算法

Kruskal 算法根据选权值最小的边进行扩张，采用类似并查集的合并方法。

#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1e5 + 10;
struct node {
     
    int x, y, len;
} edge[maxn << 1];
int n, k, w, cnt = 0;
char u, v;
int fa[maxn];

void add_edge(int x, int y, int w) {
     
    edge[cnt].len = w;
    edge[cnt].x = y;
    edge[cnt].y = x;
    cnt++;
}
bool cmp(node a, node b) {
     
    return a.len < b.len;
}

int find_fa(int x) {
     
    return fa[x] = (fa[x] == x?x:find_fa(fa[x]));
}

bool join(int x, int y) {
     
    if (find_fa(x) == find_fa(y)) return false;
    fa[find_fa(x)] = find_fa(y);
    return true;
}

int kruskal() {
     
    int ans = 0;
    for (int i = 0; i < cnt; i++) {
     
        if (join(edge[i].x, edge[i].y)) {
     
            ans += edge[i].len;
        }
    }
    return ans;
}


int main()
{
     
    while (cin >> n && n) {
     
        for (int i = 0; i < maxn; i++) {
     
            fa[i] = i;
        }
        cnt = 0;
        for (int i = 0; i < n - 1; i++) {
     
            cin >> u >> k;
            for (int j = 0; j < k; j++) {
     
                cin >> v >> w;
                add_edge(u - 'A', v - 'A', w);
            }
        }
        sort(edge, edge + cnt, cmp);

        printf("%d\n", kruskal());
    }

    return 0;
}