裸Dijkstra算法

Til the Cows Come Home
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 47476 Accepted: 16144

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. 

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it. 

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N 
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>
#include<queue>;
#include<stack>;
#include <iomanip>;
#define INF 0x3f3f3f3f
using namespace std;
int a[2100][2100];
int d[2100];
bool v[2100];
int main()
{
    int t,n,i,j,x1,x2,x3;
    while(~scanf("%d%d",&t,&n))
    {
        memset(v,0,sizeof(v));
        for(i=1; i<=2000; i++)
        {
            for(j=1; j<=2000; j++)
                a[i][j]=INF;
        }
        while(t--)
        {
            scanf("%d%d%d",&x1,&x2,&x3);
            if(a[x1][x2]>x3)
                a[x2][x1]=a[x1][x2]=x3;
        }
        for(int i=1; i<=n; i++)
        {
            d[i]=a[1][i];
        }
        d[1]=0;
        for(i=1; i<=n; i++)
        {
            int x,m=INF;
            for( j=1; j<=n; j++)
            {
                if(!v[j]&&d[j]<m)
                    m=d[x=j];
            }
            v[x]=1;
            for( j=1; j<=n; j++)
            {
                d[j]=min(d[j],d[x]+a[x][j]);
            }
        }
        printf("%d\n",d[n]);
    }
}


* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

Sample Output

90

Hint

INPUT DETAILS: 

There are five landmarks. 

OUTPUT DETAILS: 

Bessie can get home by following trails 4, 3, 2, and 1.

Source

题意:就是求从n到1的最短路 
(Kriging_NSGA2)克里金模型结合多目标遗传算法求最优因变量及对应的最佳自变量组合研究(Matlab代码实现)内容概要:本文介绍了克里金模型(Kriging)与多目标遗传算法NSGA-II相结合的方法,用于求解最优因变量及其对应的最佳自变量组合,并提供了完整的Matlab代码实现。该方法首先利用克里金模型构建高精度的代理模型,逼近复杂的非线性系统响应,减少计算成本;随后结合NSGA-II算法进行多目标优化,搜索帕累托前沿解集,从而获得多个最优折衷方案。文中详细阐述了代理模型构建、算法集成流程及参数设置,适用于工程设计、参数反演等复杂优化问题。此外,文档还展示了该方法在SCI一区论文中的复现应用,体现了其科学性与实用性。; 适合人群:具备一定Matlab编程基础,熟悉优化算法和数值建模的研究生、科研人员及工程技术人员,尤其适合从事仿真优化、实验设计、代理模型研究的相关领域工作者。; 使用场景及目标:①解决高计算成本的多目标优化问题,通过代理模型降低仿真次数;②在无法解析求导或函数高度非线性的情况下寻找最优变量组合;③复现SCI高水平论文中的优化方法,提升科研可信度与效率;④应用于工程设计、能源系统调度、智能制造等需参数优化的实际场景。; 阅读建议:建议读者结合提供的Matlab代码逐段理解算法实现过程,重点关注克里金模型的构建步骤与NSGA-II的集成方式,建议自行调整测试函数或实际案例验证算法性能,并配合YALMIP等工具包扩展优化求解能力。
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