1、题意:dijkstra模板题,存点模板

#include <queue>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 2000010
#define inf 1047483647

inline int read(){
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9'){
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while('0' <= ch && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

namespace dijkstra{
    struct Node{
        int d, u;
        inline bool operator < (const Node& rhs) const{
            return d > rhs.d;
        }
    };
    int d[M], done[M];
    priority_queue<Node> Q;
    struct Edge{
        int u, v, w, next;
    } G[M];
    int head[M], tot;
    int n; //number of points

    inline void init(){
        memset(head, -1, sizeof(head));
        tot = -1;
    }

    inline void add(int u, int v, int w){
        G[++ tot] = (Edge){u, v, w, head[u]};
        head[u] = tot;
    }

    inline void get_dis(int s){
        for(int i = 1; i <= n; i ++) d[i] = inf;
        d[s] = 0; while(!Q.empty()) Q.pop();
        Q.push((Node){0, s});
        memset(done, 0, sizeof(done));
        while(!Q.empty()){
            Node u = Q.top(); Q.pop();
            int x = u.u;
            if(done[x]) continue;
            done[x] = 1;
            for(int i = head[x]; i != -1; i = G[i].next){
                Edge& e = G[i];
                if(d[e.v] > d[x] + e.w){
                    d[e.v] = d[x] + e.w;
                    Q.push((Node){d[e.v], e.v});
                }
            }
        }
    }
}

using namespace dijkstra;

int main(){
    n = read();
    int m = read();
    init();
    for(int i = 1; i <= m; i ++){
        int u = read(), v = read(), w = read();
        add(u, v, w);
    }
    get_dis(1);
    printf("%d\n", d[n]);
    return 0;
}

BZOJ2292——【POJ Challenge 】永远挑战的更多相关文章

  1. bzoj2292【POJ Challenge 】永远挑战*

    bzoj2292[POJ Challenge ]永远挑战 题意: 有向图,每条边长度为1或2,求1到n最短路.点数≤100000,边数≤1000000. 题解: 有人说spfa会T,所以我用了dijk ...

  2. BZOJ2292: 【POJ Challenge 】永远挑战

    2292: [POJ Challenge ]永远挑战 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 513  Solved: 201[Submit][ ...

  3. 2292: 【POJ Challenge 】永远挑战

    2292: [POJ Challenge ]永远挑战 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 553  Solved: 230[Submit][ ...

  4. bzoj 2295: 【POJ Challenge】我爱你啊

    2295: [POJ Challenge]我爱你啊 Time Limit: 1 Sec  Memory Limit: 128 MB Description ftiasch是个十分受女生欢迎的同学,所以 ...

  5. 【链表】BZOJ 2288: 【POJ Challenge】生日礼物

    2288: [POJ Challenge]生日礼物 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 382  Solved: 111[Submit][S ...

  6. BZOJ2288: 【POJ Challenge】生日礼物

    2288: [POJ Challenge]生日礼物 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 284  Solved: 82[Submit][St ...

  7. BZOJ2293: 【POJ Challenge】吉他英雄

    2293: [POJ Challenge]吉他英雄 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 80  Solved: 59[Submit][Stat ...

  8. BZOJ2287: 【POJ Challenge】消失之物

    2287: [POJ Challenge]消失之物 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 254  Solved: 140[Submit][S ...

  9. BZOJ2295: 【POJ Challenge】我爱你啊

    2295: [POJ Challenge]我爱你啊 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 126  Solved: 90[Submit][Sta ...

  10. BZOJ2296: 【POJ Challenge】随机种子

    2296: [POJ Challenge]随机种子 Time Limit: 1 Sec  Memory Limit: 128 MBSec  Special JudgeSubmit: 114  Solv ...

随机推荐

  1. [LeetCode] Implement Queue using Stacks 用栈来实现队列

    Implement the following operations of a queue using stacks. push(x) -- Push element x to the back of ...

  2. Ubuntu 16.04 LAMP server 指南 - 配置 Apache2.4,PHP7,和MariaDB(而不是MySQL)

    翻译自:https://www.howtoforge.com/tutorial/install-apache-with-php-and-mysql-on-ubuntu-16-04-lamp/ 昨天在虚 ...

  3. 【MySQL】mysql workbench

    1.导入,导出,数据之间的传输[两台服务器]2.连接管理3.服务器管理4.表的管理

  4. AngularJS模型

    1. AngularJS模型主要就是使用的AngularJS的ng-model指令. ng-model指令可以将输入域的值与 AngularJS 创建的变量绑定. <!DOCTYPE html& ...

  5. React JS的基本用法[ES5,纯前端写法]

    1.配置webpack npm install -g webpack #webpack的cli npm install -g webpack-dev-server #webpack自带的服务器 npm ...

  6. Codeforces #364 DIV2

      ~A题 A. Cards time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  7. lua命令行编译

    http://jingyan.baidu.com/article/359911f551917457fe0306e5.html 最后将生成的.exe解释器的根目录配置到系统环境变量 copy lua.c ...

  8. mysql Packet for query is too large (1185 > 1024)异常

    注:最近mysql一直提示如下错误 Packet for query is too large (1185 > 1024). You can change this value on the s ...

  9. 几个MQTT的知识点

    开始正文前需要感谢一下网友“小龙”和emqtt.io群里的网友们的帮助,本人刚刚开始使用MQTT有很多不懂的地方,在emqtt.io群里询问解决方法的时候,“小龙”给我详细的讲解了一些MQTT的知识点 ...

  10. Mysql两个引擎对比

    Mysql两个引擎对比 MyIsam      优点:      1.支持B-Tree检索和文本全文检索      2.性能消耗方面相对较低      3.支持全表(table)锁      缺点: ...