hdu 3371 Connect the Cities

最新推荐文章于 2025-12-31 07:08:05 发布

转载最新推荐文章于 2025-12-31 07:08:05 发布 · 93 阅读

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原文链接：http://www.cnblogs.com/GadyPu/p/4792917.html

文章标签：

#c/c++ #php

本文详细解析了一道关于最小生成树算法的编程题，通过Kruskal算法实现城市间的道路连接，以最少的成本使所有城市相连。介绍了算法的实现过程，包括数据结构定义、关键函数如初始化、查找和合并操作，以及具体的输入输出处理。

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=3371

Connect the Cities

Description

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.

Input

The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

Sample Input

1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6

Sample Output

最小生成树，这道题卡时间g++过了，c++ tle了/(ㄒoㄒ)/~~。。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
using std::map;
using std::min;
using std::sort;
using std::pair;
using std::vector;
using std::multimap;
using std::priority_queue;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 510;
const int INF = 0x3f3f3f3f;
struct edge {
    int u, v, w;
    inline bool operator<(const edge &x) const {
        return w < x.w;
    }
}G[(N * N) << 1];
struct Kruskal {
    int E, par[N], rank[N];
    inline void init(int n) {
        E = 0;
        rep(i, n + 2) {
             par[i] = i;
             rank[i] = 0;
        }
    }
    inline int find(int x) {
        while(x != par[x]) {
            x = par[x] = par[par[x]];
        }
        return x;
    }
    inline bool unite(int x, int y) {
        x = find(x), y = find(y);
        if(x == y) return false;
        if(rank[x] < rank[y]) {
            par[x] = y;
        } else {
            par[y] = x;
            rank[x] += rank[x] == rank[y];
        }
        return true;
    }
    inline void built(int m, int k) {
        int u, v, w, q, fa;
        while(m--) {
            scanf("%d %d %d", &u, &v, &w);
            G[E++] = { u, v, w };
        }
        while(k--) {
            scanf("%d", &q);
            scanf("%d %d", &u, &v);
            G[E++] = { u, v, 0 };
            fa = v;
            rep(i, q - 2) {
                scanf("%d", &v);
                G[E++] = { fa, v, 0 };
                fa = v;
            }
        }
    }
    inline int kruskal(int n) {
        int ans = 0, cnt = 0;
        sort(G, G + E);
        rep(i, E) {
            int u = G[i].u, v = G[i].v;
            if(unite(u, v)) {
                ans += G[i].w;
                if(++cnt >= n - 1) return ans;
            }
        }
        return -1;
    }
    inline void solve(int n, int m, int k) {
        init(n), built(m, k);
        printf("%d\n", kruskal(n));
    }
}go;
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w+", stdout);
#endif
    int n, m, k, T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d %d %d", &n, &m, &k);
        go.solve(n, m, k);
    }
    return 0;
}

转载于:https://www.cnblogs.com/GadyPu/p/4792917.html