zoukankan      html  css  js  c++  java
  • luogu P4009 汽车加油行驶问题

    网络流(×),多层图BFS(√)

    typedef long long LL;
    typedef pair<LL, LL> PLL;
    typedef pair<int, int> PINT;
    
    const int maxm = 115;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int dx[] = {1, -1, 0, 0};
    const int dy[] = {0, 0, 1, -1};
    
    LL d[maxm][maxm][13];
    int G[maxm][maxm];
    bool inq[maxm][maxm][13];
    
    struct Node {
        int x, y, k;
    };
    
    void run_case() {
        memset(d, 63, sizeof(d));
        int N, K, A, B, C;
        cin >> N >> K >> A >> B >> C;
        for(int i = 1; i <= N; ++i)
            for(int j = 1; j <= N; ++j)
                cin >> G[i][j];
        queue<Node> q;
        q.push(Node{1, 1, K}); inq[1][1][K] = true;
        d[1][1][K] = 0;
        while(!q.empty()) {
            auto now = q.front(); q.pop();
            inq[now.x][now.y][now.k] = false;
            if(G[now.x][now.y] && now.k != K) {
                if(d[now.x][now.y][K] > d[now.x][now.y][now.k]+A) {
                    d[now.x][now.y][K] = d[now.x][now.y][now.k]+A;
                    if(!inq[now.x][now.y][K]) {
                        inq[now.x][now.y][K] = true;
                        q.push(Node{now.x, now.y, K});
                    }
                }
                continue;
            } else if(!G[now.x][now.y]){
                if(d[now.x][now.y][K] > d[now.x][now.y][now.k]+A+C) {
                    d[now.x][now.y][K] = d[now.x][now.y][now.k]+A+C;
                    if(!inq[now.x][now.y][K]) {
                        inq[now.x][now.y][K] = true;
                        q.push(Node{now.x, now.y, K});
                    }
                }
            }
            if(now.k > 0) {
                for(int i = 0; i < 4; ++i) {
                    int nx = now.x + dx[i], ny = now.y + dy[i];
                    if(nx < 1 || nx > N || ny < 1 || ny > N) continue;
                    int cost = 0;
                    if(nx < now.x || ny < now.y) cost = B;
                    if(d[nx][ny][now.k-1] > d[now.x][now.y][now.k] + cost) {
                        d[nx][ny][now.k-1] = d[now.x][now.y][now.k] + cost;
                        if(!inq[nx][ny][now.k-1]) {
                            inq[nx][ny][now.k-1] = true;
                            q.push(Node{nx, ny, now.k-1});
                        }
                    }
                }
            }
            
        }
        LL ans = INF+5;
        for(int i = 0; i <= K; ++i)
            ans = min(ans, d[N][N][i]);
        cout << ans;
    }
    
    int main() {
        ios::sync_with_stdio(false), cin.tie(0);
        run_case();
        //cout.flush();
        return 0;
    }
    View Code
  • 相关阅读:
    Delphi / C++ Builder 使用 UDT ( UDP-based Data Transfer ) 4.11
    STUN: NAT 类型检测方法
    udt nat traverse
    UDT: Breaking the Data Transfer Bottleneck
    Freescale OSBDM JM60仿真器
    How To: Perl TCP / UDP Socket Programming using IO::Socket::INET
    NAT类型与穿透 及 STUN TURN 协议
    根据PID和VID得到USB转串口的串口号
    pic/at89c2051 programmer
    IC开短路测试(open_short_test),编程器测试接触不良、开短路
  • 原文地址:https://www.cnblogs.com/GRedComeT/p/12283367.html
Copyright © 2011-2022 走看看