zoukankan      html  css  js  c++  java
  • hdu 2988 Dark roads

    题目连接

    http://acm.hdu.edu.cn/showproblem.php?pid=2988  

    Dark roads

    Description

    Economic times these days are tough, even in Byteland. To reduce the operating costs, the government of Byteland has decided to optimize the road lighting. Till now every road was illuminated all night long, which costs 1 Bytelandian Dollar per meter and day. To save money, they decided to no longer illuminate every road, but to switch off the road lighting of some streets. To make sure that the inhabitants of Byteland still feel safe, they want to optimize the lighting in such a way, that after darkening some streets at night, there will still be at least one illuminated path from every junction in Byteland to every other junction. 

    What is the maximum daily amount of money the government of Byteland can save, without making their inhabitants feel unsafe? 

    Input

    The input file contains several test cases. Each test case starts with two numbers m and n, the number of junctions in Byteland and the number of roads in Byteland, respectively. Input is terminated by m=n=0. Otherwise, 1 ≤ m ≤ 200000 and m-1 ≤ n ≤ 200000. Then follow n integer triples x, y, z specifying that there will be a bidirectional road between x and y with length z meters (0 ≤ x, y < m and x ≠ y). The graph specified by each test case is connected. The total length of all roads in each test case is less than 231.

    Output

    For each test case print one line containing the maximum daily amount the government can save. 

    Sample Input

    7 11
    0 1 7
    0 3 5
    1 2 8
    1 3 9
    1 4 7
    2 4 5
    3 4 15
    3 5 6
    4 5 8
    4 6 9
    5 6 11
    0 0

    Sample Output

    51

    最小生成树。。

    #include<algorithm>
    #include<iostream>
    #include<cstdlib>
    #include<cstring>
    #include<cstdio>
    #include<vector>
    #include<queue>
    #include<set>
    using std::set;
    using std::sort;
    using std::pair;
    using std::swap;
    using std::multiset;
    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 = 200010;
    const int INF = 0x3f3f3f3f;
    typedef unsigned long long ull;
    struct edge {
        int u, v, w;
        inline bool operator<(const edge &x) const {
            return w < x.w;
        }
    }G[N];
    struct Kruskal {
        int E, sum, par[N], rank[N];
        inline void init(int n) {
            E = sum = 0;
            rep(i, n + 1) {
                par[i] = i, rank[i] = 0;
            }
        }
        inline void built(int m) {
            int u, v, w;
            while (m--) {
                scanf("%d %d %d", &u, &v, &w);
                G[E++] = { u, v, w }, sum += w;
            }
        }
        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 int kruskal() {
            int ans = 0;
            sort(G, G + E);
            rep(i, E) {
                edge &e = G[i];
                if (unite(e.u, e.v)) {
                    ans += e.w;
                }
            }
            return ans;
        }
        inline void solve(int n, int m) {
            init(n), built(m);
            printf("%d
    ", sum - kruskal());
        }
    }go;
    int main() {
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w+", stdout);
    #endif
        int n, m;
        while (~scanf("%d %d", &n, &m), n + m) {
            go.solve(n, m);
        }
        return 0;
    }
  • 相关阅读:
    HDU 1025:Constructing Roads In JGShining's Kingdom(LIS+二分优化)
    HDU 3938:Portal(并查集+离线处理)
    HDU 1811:Rank of Tetris(并查集+拓扑排序)
    HDU 1074:Doing Homework(状压DP)
    HDU 1024:Max Sum Plus Plus(DP)
    最最最亲爱哒
    hlg-1332 买电脑 ---二分
    时间过得很快
    0514
    hlg1551Assemble--暴力求解
  • 原文地址:https://www.cnblogs.com/GadyPu/p/4783980.html
Copyright © 2011-2022 走看看