HDU 2544 最短路

HDU 2544 最短路

Problem Description

在每年的校賽裏，所有進入決賽的同學都會獲得一件很漂亮的t-shirt。但是每當我們的工作人員把上百件的衣服從商店運回到賽場的時候，卻是非常累的！所以現在他們想要尋找最短的從商店到賽場的路線，你可以幫助他們嗎？

Input

輸入包括多組數據。每組數據第一行是兩個整數N、M（N<=100，M<=10000），N表示成都的大街上有幾個路口，標號爲1的路口是商店所在地，標號爲N的路口是賽場所在地，M則表示在成都有幾條路。N=M=0表示輸入結束。接下來M行，每行包括3個整數A，B，C（1<=A,B&lt;=N,1<=C<=1000）,表示在路口A與路口B之間有一條路，我們的工作人員需要C分鐘的時間走過這條路。
輸入保證至少存在1條商店到賽場的路線。

Output

對於每組輸入，輸出一行，表示工作人員從商店走到賽場的最短時間

Sample Input

2 1 1 2 3 3 3 1 2 5 2 3 5 3 1 2 0 0

Sample Output

3 2

Source

UESTC 6th Programming Contest Online

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lcy

Solution

三種解法（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 long
using 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 long
using 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 long
using 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

多組數據需要注意預處理