[最短路径] HDU 1690 - Bus System

来源：互联网发布：上海达观数据公司咋样编辑：程序博客网时间：2024/06/15 03:56

给定车辆的收费标准与路程区间的关系如下：
路程区间花费 (0, L₁] C₁ (L₁, L₂] C₂ (L₂, L₃] C₃ (L₃, L₄] C₄ (L₄, +∞] 没票
接着输入两个整数N, M。
接着输入N个车站所在的横坐标，这N个车站是在一条直线上的。
接下来有M个询问，S_i, E_i分别代表起点和终点。
求起点到终点的最少花费?

由于N的范围比较小, 可以把这些车站两两之间距离存储到邻接矩阵中，然后跑一下最短路径即可。
注意INF的初值要大一点。

#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <algorithm>#include <iostream>#include <set>#include <map>#include <queue>#include <stack>#include <assert.h>#include <time.h>typedef long long LL;const LL INF = 1e18;const double EPS = 1e-9;const double PI = acos(-1.0);using namespace std;LL dis[105], graph[105][105];int N, vis[105];void init(){    for(int i = 0; i <= 100; i++)    {        for(int j = 0; j <= 100; j++)        {            graph[i][j] = INF;        }    }}int abs(int x){    if(x < 0) return -x;    return x;}void Dijkstra(int s){    for(int i = 1; i <= N; i++)    {        dis[i] = graph[s][i];        vis[i] = -1;    }    vis[s] = 0;    for(int i = 2; i <= N; i++)    {        LL minn = INF, k = -1;        for(int j = 1; j <= N; j++)        {            if(vis[j] == -1 && dis[j] < minn)            {                minn = dis[j];                k = j;            }        }        if(k == -1) break;        vis[k] = 0;        for(int j = 1; j <= N; j++)        {            if(vis[j] == -1 && dis[j] > dis[k] + graph[k][j])            {                dis[j] = dis[k] + graph[k][j];            }        }    }}int main(){    #ifdef _T1est        freopen("test0.in", "r", stdin);        freopen("test0.out", "w", stdout);        srand(time(NULL));    #endif    int T, L[10], C[10], M, val[105];    scanf("%d", &T);    for(int i = 1; i <= T; i++)    {        printf("Case %d:\n", i);        init();        for(int ii = 0; ii < 4; ii++)        {            scanf("%d", &L[ii]);        }        for(int ii = 0; ii < 4; ii++)        {            scanf("%d", &C[ii]);        }        scanf("%d %d", &N, &M);        for(int ii = 1; ii <= N; ii++)        {            scanf("%d", &val[ii]);        }        for(int ii = 1; ii <= N; ii++)        {            for(int jj = 1; jj <= N; jj++)            {                if(ii != jj)                {                    int tmp = abs(val[ii] - val[jj]);                    for(int k = 0; k < 4; k++)                    {                        if(tmp <= L[k])                        {                            graph[ii][jj] = C[k];                            break;                        }                    }                }            }        }        for(int ii = 0; ii < M; ii++)        {            int s, e;            scanf("%d %d", &s, &e);            Dijkstra(s);            if(dis[e] == INF)            {                printf("Station %d and station %d are not attainable.", s, e);            }            else            {                printf("The minimum cost between station %d and station %d is %I64d.", s, e, dis[e]);            }            puts("");        }    }    return 0;}

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