hdu 1690 Bus System ---Floyd

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这道题用Floyd更方便。。

复制代码#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; }

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#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;
}