hdu 1690 Bus System ---Floyd

最新推荐文章于 2020-04-07 23:53:58 发布

Rki-dor

最新推荐文章于 2020-04-07 23:53:58 发布

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原文链接：http://www.cnblogs.com/yyf573462811/archive/2012/07/25/6365390.html

本文介绍了一个使用Floyd算法解决特定路径查找问题的实例。通过详细解释代码实现过程，包括初始化地图矩阵、输入处理、Floyd算法的具体应用以及输出结果等步骤，展示了如何有效地利用Floyd算法来计算任意两个节点之间的最短路径。

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这道题用Floyd更方便。。

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <cmath>
using namespace std;

const long long INF = 99999999999LL;    //注意！要加LL，不然会报错数据太大
const int N = 105;

int l1, l2, l3, l4, c1, c2, c3, c4;
long long map[N][N];    //距离可能会爆int，所以用long long
int place[N];
int n, m;

void init()
{
    int i, j;
    for(i = 1; i <= n; i++)
        for(j = 1; j <= n; j++)
            if(i == j) map[i][j] = 0;
            else map[i][j] = INF;
}

void input()
{
    int i, j, len;
    scanf("%d%d%d%d%d%d%d%d", &l1, &l2, &l3, &l4, &c1, &c2, &c3, &c4);
    scanf("%d %d", &n, &m);
    init();
    for(i = 1; i <= n; i++)
    {
        scanf("%d", &place[i]);
    }
    for(i = 1; i <= n; i++)
    {
        for(j = i+1; j <= n; j++)
        {
            len = abs(place[i] - place[j]);
            if(0 < len && len <= l1) map[i][j] = map[j][i] = c1;
            else if(l1 < len && len <= l2) map[i][j] = map[j][i] = c2;
            else if(l2 < len && len <= l3) map[i][j] = map[j][i] = c3;
            else if(l3 < len && len <= l4) map[i][j] = map[j][i] = c4;
        }
    }
}

void floyd()    //这题绝对是用floyd方便
{
    int i, j, k;
    for(k = 1; k <= n; k++)
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
                if(map[i][j] > map[i][k] + map[k][j])
                     map[i][j] = map[i][k] + map[k][j];
}

void output()
{
    int ti, tj;
    static int zz = 1;
    printf("Case %d:\n", zz++);
    while(m--)
    {
        scanf("%d %d", &ti, &tj);
        if(map[ti][tj] != INF)
            printf("The minimum cost between station %d and station %d is %I64d.\n", ti, tj, map[ti][tj]);
        else
            printf("Station %d and station %d are not attainable.\n", ti, tj);
    }
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        input();
        floyd();
        output();
    }

    return 0;
}

转载于:https://www.cnblogs.com/yyf573462811/archive/2012/07/25/6365390.html