[最短路径] HDU 1690 - Bus System

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，Si, Ei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;
}