uva 10801 - Lift Hopping(最短路Dijkstra)

题目链接：10801 - Lift Hopping

题目大意：有一栋100层的大楼（标号为0~99），里面有n个电梯（不超过5个），以及要到达的层数（aid)，然后是每个电梯走一层所需的时间，再n行就是对应每个电梯可以到达的层数，数量不定。然后每装换一次电梯需要等待60秒，问，最快能多快到达目标层数。

解题思路：这题有点坑啊，一开始是用邻接表+Dijkstra，可是忘记考虑aid = 0的情况，这种情况下所得到的答案不需要减掉60（为了方便计算，每次到达一个楼层时间就+60），可是我没发现这个问题，就换成用邻接矩阵去做，后来发现了aid = 0的情况，可是用邻接矩阵需要注意的是更新g[a][b]的为最小值，即当两个电梯都能满足从a层到b层的时候，要选取小的值做g[a][b]的值。

1.邻接矩阵

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
const int N = 105;
const int INF = 0x3f3f3f3f;
int n, aid, cnt, time
, d
;
int g

;

void add(int a, int b, int s) {
int dis = abs(b - a) * s;
if (g[a][b] > dis)
g[a][b] = g[b][a] = dis;
}

void init() {
for (int i = 0; i < n; i++)
scanf("%d", &time[i]);

memset(d, 0x3f, sizeof(d));
memset(g, 0x3f, sizeof(g));
d[0] = 0;

for (int i = 0; i < n; i++) {
char ch = '\0';
int num
;
for (int j = 0; ch != '\n'; j++) {
scanf("%d%c", &num[j], &ch);
for (int k = 0; k < j; k++)
add(num[j], num[k], time[i]);
}
}
}

void solve() {
int vis
;
memset(vis, 0, sizeof(vis));
for (int i = 0; i < 99; i++) {
int x, m = INF, flag = 0;
for (int j = 0; j < 100; j++)
if (!vis[j] && d[j] < m) { m = d[j], x = j, flag = 1; }
if (!flag) return;
vis[x] = 1;
for (int j = 0; j < 100; j++) {
if (!vis[j] && d[j] > d[x] + g[x][j] + 60)
d[j] = d[x] + g[x][j] + 60;
}
}
}

int main () {
while (scanf("%d%d", &n, &aid) == 2) {
init();
solve();
if (aid == 0) printf("0\n");
else if (d[aid] != INF) printf("%d\n", d[aid] - 60);
else printf("IMPOSSIBLE\n");
}
return 0;
}

2.邻接表

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
const int N = 105;
const int M = 500005;
const int INF = 1 << 30;
int n, aid, cnt, time
, d
;
int first[M], next[M], u[M], v[M], w[M];

void add(int a, int b, int s) {
u[cnt] = a;
v[cnt] = b;
w[cnt] = abs(b - a) * s;
next[cnt] = first[u[cnt]];
first[u[cnt]] = cnt;
cnt++;
}

void init() {
for (int i = 0; i < n; i++)
scanf("%d", &time[i]);

for (int i = 0; i < 100; i++) {
first[i] = -1;
d[i] = (i == 0 ? 0 : INF);
}

for (int i = 0; i < n; i++) {
char ch = '\0';
int num
;
for (int j = 0; ch != '\n'; j++) {
scanf("%d%c", &num[j], &ch);
for (int k = 0; k < j; k++)
add(num[j], num[k], time[i]), add(num[k], num[j], time[i]);
}
}
}

void solve() {
int vis
;
memset(vis, 0, sizeof(vis));
for (int i = 0; i < 100; i++) {
int x, m = INF;
for (int j = 0; j < 100; j++)
if (!vis[j] && d[j] < m) { m = d[j], x = j; }
vis[x] = 1;
for (int j = first[x]; j != -1; j = next[j]) {
if (d[x] < INF && d[v[j]] > d[x] + w[j] + 60)
d[v[j]] = d[x] + w[j] + 60;
}
}
}

int main () {
while (scanf("%d%d", &n, &aid) == 2) {
init();
solve();
if (aid == 0) printf("0\n");
else if (d[aid] != INF) printf("%d\n", d[aid] - 60);
else printf("IMPOSSIBLE\n");
}
return 0;
}