• Connect the Cities[HDU3371]


    Connect the Cities

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 18322 Accepted Submission(s): 4482


    Problem 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
    1

    kruskal会超时,要用prim。朴素的prim也是极限时间过的,最好加个堆优化。

    #include <stdio.h>
    #include <string.h>
    using namespace std;
    class Prim {
    #define Prim_MAXN 505
    #define Prim_MAXM 100005
    public:
        int N, M;
        int head[Prim_MAXN], dis[Prim_MAXN];
        bool vis[Prim_MAXN];
        struct EDGE {
            int v, d, nex;
        } edge[Prim_MAXM];
        Prim() {
            clear();
        }
        void clear() {
            N = M = 0;
            memset(head, -1, sizeof(head));
        }
        void addEdge(int a, int b, int c) {
            edge[M].v = b;
            edge[M].d = c;
            edge[M].nex = head[a];
            head[a] = M++;
            edge[M].v = a;
            edge[M].d = c;
            edge[M].nex = head[b];
            head[b] = M++;
        }
        int MST() {
            int ret = 0, last, next, min;
            for (int i = 1; i <= N; i++) {
                dis[i] = 0x7FFFFFFF;
            }
            memset(vis, false, sizeof(vis));
            vis[1] = true;
            last = 1;
            for (int i = 1; i < N; i++) {
                for (int e = head[last]; e != -1; e = edge[e].nex) {
                    if (dis[edge[e].v] > edge[e].d) {
                        dis[edge[e].v] = edge[e].d;
                    }
                }
                min = 0x7FFFFFFF;
                for (int j = 1; j <= N; j++) {
                    if (dis[j] < min && !vis[j]) {
                        min = dis[j];
                        next = j;
                    }
                }
                if (min == 0x7FFFFFFF) {
                    return -1;
                }
                vis[next] = true;
                ret += dis[next];
                last = next;
            }
            return ret;
        }
    };
    Prim Pr;
    int main() {
        int n, m, t, k;
        scanf("%d", &t);
        while (t--) {
            Pr.clear();
            scanf("%d%d%d", &n, &m, &k);
            Pr.N = n;
            for (int i = 0; i < m; i++) {
                int a, b, c;
                scanf("%d%d%d", &a, &b, &c);
                Pr.addEdge(a, b, c);
            }
            int nn, x[505];
            for (int i = 0; i < k; i++) {
                scanf("%d", &nn);
                for (int j = 0; j < nn; j++) {
                    scanf("%d", &x[j]);
                }
                for (int j = 1; j < nn; j++) {
                    Pr.addEdge(x[0], x[j], 0);
                }
            }
            printf("%d
    ", Pr.MST());
        }
        return 0;
    }
    View Code

     

  • 相关阅读:
    一些无意间YY的脑瘫题
    回滚莫队
    数论
    专题整理
    模拟赛x+1
    HISKrrr的板子库
    java多线程学习笔记(四)
    java多线程学习笔记(三)
    java多线程学习笔记(二)
    Java多线程学习笔记(一)
  • 原文地址:https://www.cnblogs.com/dramstadt/p/6211452.html
Copyright © 2020-2023  润新知