HDU 2544 最短路

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HDU 2544 最短路

Problem Description

在每年的校赛里，所有进入决赛的同学都会获得一件很漂亮的t-shirt。但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候，却是非常累的！所以现在他们想要寻找最短的从商店到赛场的路线，你可以帮助他们吗？

Input

输入包括多组数据。每组数据第一行是两个整数N、M（N<=100，M<=10000），N表示成都的大街上有几个路口，标号为1的路口是商店所在地，标号为N的路口是赛场所在地，M则表示在成都有几条路。N=M=0表示输入结束。接下来M行，每行包括3个整数A，B，C（1<=A,B<=N,1<=C<=1000）,表示在路口A与路口B之间有一条路，我们的工作人员需要C分钟的时间走过这条路。
输入保证至少存在1条商店到赛场的路线。

Output

对于每组输入，输出一行，表示工作人员从商店走到赛场的最短时间

Sample Input

2 11 2 33 31 2 52 3 53 1 20 0

Sample Output

Source

UESTC 6th Programming Contest Online

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lcy

Solution

三种解法（Floyed、Dijkstra、SPFA）求最短路

Code

１、Floyed：O(n^3)

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <map>#include <vector>#include <queue>#define L 110#define inf 1000000009#define LL long longusing namespace std;int n, m, d[L][L], a, b, c, dis[L];bool vis[L];inline void Floyed() {  for (int k = 1; k <= n; ++k)    for (int i = 1; i <= n; ++i)      if (d[i][k] != inf)for (int j = 1; j <= n; ++j)  if (d[i][j] > d[i][k] + d[j][k]) d[i][j] = d[j][i] = d[i][k] + d[j][k];  printf("%d\n", d[1][n]);}int main(){  freopen("HDU2544.in", "r", stdin);  freopen("HDU2544.out", "w", stdout);  while (scanf("%d %d", &n, &m) != EOF && n && m) {    for (int i = 1; i <= n; ++i)      for (int j = 1; j <= n; ++j) d[i][j] = inf;    for (int i = 1; i <= m; ++i)      scanf("%d %d %d", &a, &b, &c), d[a][b] = d[b][a] = c;    Floyed();  }  return 0;}

2、Dijkstra：O(n^2)

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <map>#include <vector>#include <queue>#define L 110#define inf 1000000009#define LL long longusing namespace std;int n, m, d[L][L], a, b, c, dis[L];bool vis[L];inline void dijkstra() {  for (int i = 1; i <= n; ++i) dis[i] = d[1][i], vis[i] = false;    vis[1] = true;    for (int i = 1; i <= n; ++i) {      int temp, minx = inf;      for (int j = 1; j <= n; ++j)if (!vis[j] && dis[j] < minx) minx = dis[j], temp = j;      vis[temp] = true;      for (int j = 1; j <= n; ++j)if (!vis[j] && d[j][temp] + dis[temp] < dis[j]) dis[j] = d[j][temp] + dis[temp];    }    printf("%d\n", dis[n]);}int main(){  freopen("HDU2544.in", "r", stdin);  freopen("HDU2544.out", "w", stdout);  while (scanf("%d %d", &n, &m) != EOF && n && m) {    for (int i = 1; i <= n; ++i)      for (int j = 1; j <= n; ++j) d[i][j] = inf;    for (int i = 1; i <= m; ++i)      scanf("%d %d %d", &a, &b, &c), d[a][b] = d[b][a] = c;    dijkstra();  }  return 0;}

3、SPFA：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <map>#include <vector>#include <queue>#define L 10010#define inf 1000000009#define LL long longusing namespace std;struct node {  int nxt, to, w;} e[L << 1];int n, m, d[L], head[L], a, b, c, cnt;bool vis[L];queue <int> q;inline void add(int a, int b, int c) {  e[++cnt].nxt = head[a], e[cnt].to = b, e[cnt].w = c, head[a] = cnt;}int main(){  freopen("HDU2544.in", "r", stdin);  freopen("HDU2544.out", "w", stdout);  while (scanf("%d %d", &n, &m) != EOF && n && m) {    memset(head, 0, sizeof(head));    memset(vis, 0, sizeof(vis));    for (int i = 2; i <= n; ++i) d[i] = inf;    d[1] = 0, cnt = 0;    for (int i = 1; i <= m; ++i)      scanf("%d %d %d", &a, &b, &c), add(a, b, c), add(b, a, c);    while(!q.empty()) q.pop();    q.push(1);    while (!q.empty()) {      int x = q.front();      q.pop();      vis[x] = 0;      for (int i = head[x]; i; i = e[i].nxt) {int y = e[i].to;if (d[y] > d[x] + e[i].w) {  d[y] = d[x] + e[i].w;  if (!vis[y]) vis[y] = 1, q.push(y);}      }    }    printf("%d\n", d[n]);  }  return 0;}

Summary

多组数据需要注意预处理

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