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  • 【2019南昌网络赛】B-Fire-Fighting Hero

    题目链接

    分析

    英雄方面很简单,跑一遍 Dijkstra 就行了,但是灭火团队就有点麻烦了。

    这里可以借助一下最大流的建边来解决这个问题:
    我们可以另外找一个点作为起点,然后建立从那个点到每一个团队的起点的边,权值为0,这样就完成了多起点的最短路

    恰好我的板子是封装好的 Dijkstra ,我就直接建立两个结构体解决问题,因为点的数量只有 1000 个,空间上已经没有什么顾虑了

    AC-Code

    #include <bits/stdc++.h>
    
    using namespace std;
    
    #define MAXN 1100
    #define MAXM 1000000
    
    #define INF 0x3fffffff              //防止后面溢出,这个不能太大
    
    struct Graph {
        struct Edge {
            long long to, next;
            long long cost;
        } edge[MAXM];
        long long head[MAXN];
        long long tot;
    
        void init(long long n) {
            tot = 0;
            memset(head, -1, sizeof(long long) * (n + 1));
        }
    
        void add_edge(long long from, long long to, long long value) {
            edge[tot].to = to;
            edge[tot].cost = value;
            edge[tot].next = head[from];
            head[from] = tot++;
        }
    };
    
    struct Dijkstra {
        long long low_cost[MAXN];
        bool vis[MAXN];
        long long pre[MAXN];
    
        void solve(long long b, long long e, long long start, Graph &graph) {
            for (long long i = b; i < e; i++) {
                low_cost[i] = INF;
                vis[i] = false;
                pre[i] = -1;
            }
            low_cost[start] = 0;
            vis[start] = true;
            long long cur_edge = graph.head[start];
            while (cur_edge != -1) {
                if (!vis[graph.edge[cur_edge].to] &&
                    low_cost[start] + graph.edge[cur_edge].cost < low_cost[graph.edge[cur_edge].to]) {
                    low_cost[graph.edge[cur_edge].to] = low_cost[start] + graph.edge[cur_edge].cost;
                    pre[graph.edge[cur_edge].to] = start;
                }
                cur_edge = graph.edge[cur_edge].next;
            }
            for (long long j = b; j < e - 1; j++) {
                long long k = -1;
                long long Min = INF;
                for (long long i = b; i < e; i++) {
                    if (!vis[i] && low_cost[i] < Min) {
                        Min = low_cost[i];
                        k = i;
                    }
                }
                if (k == -1)
                    break;
                vis[k] = true;
                cur_edge = graph.head[k];
                while (cur_edge != -1) {
                    if (!vis[graph.edge[cur_edge].to] &&
                        low_cost[k] + graph.edge[cur_edge].cost < low_cost[graph.edge[cur_edge].to]) {
                        low_cost[graph.edge[cur_edge].to] = low_cost[k] + graph.edge[cur_edge].cost;
                        pre[graph.edge[cur_edge].to] = k;
                    }
                    cur_edge = graph.edge[cur_edge].next;
                }
            }
        }
    };
    
    Graph graph;
    Dijkstra dijkstra1, dijkstra2;
    int k_node[MAXN];
    
    void solve() {
        long long t;
        cin >> t;
        long long v, e, s, k, c;
        for (int ts = 0; ts < t; ++ts) {
            cin >> v >> e >> s >> k >> c;
            graph.init(v + 1);
            for (int i = 0; i < k; ++i) {
                cin >> k_node[i];
            }
            long long from, to, value;
            for (long long i = 0; i < e; ++i) {
                cin >> from >> to >> value;
                graph.add_edge(from, to, value);
                graph.add_edge(to, from, value);
            }
            dijkstra1.solve(1, v + 1, s, graph);//第一次跑dijkstra
            for (int i = 0; i < k; ++i) {
                graph.add_edge(0, k_node[i], 0); // 这里设定超级源点为0,建立从0到每一个团队起点的边,权值为0
            }
            dijkstra2.solve(0, v + 1, 0, graph);//第二次跑dijkstra
    
            long long s_min_max = 0;
            for (long long i = 1; i < v + 1; ++i)
                s_min_max = max(s_min_max, dijkstra1.low_cost[i]);
    
            long long k_min_max = 0;
            for (long long i = 1; i < v + 1; ++i)
                k_min_max = max(k_min_max, dijkstra2.low_cost[i]);
    
            if (s_min_max <= c * k_min_max)//考虑到精度问题,这里用乘法代替
                cout << s_min_max << endl;
            else
                cout << k_min_max << endl;
        }
    }
    
    int main() {
        ios_base::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
    #ifdef ACM_LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        long long test_index_for_debug = 1;
        char acm_local_for_debug;
        while (cin >> acm_local_for_debug) {
            cin.putback(acm_local_for_debug);
            if (test_index_for_debug > 100) {
                throw runtime_error("Check the stdin!!!");
            }
            auto start_clock_for_debug = clock();
            cout << "Test " << test_index_for_debug << ":" << endl;
            solve();
            auto end_clock_for_debug = clock();
            cerr << "Test " << test_index_for_debug++ << " Run Time: "
                 << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
            cout << "
    --------------------------------------------------" << endl;
        }
    #else
        solve();
    #endif
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mauve-hkq/p/12173142.html
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