poj2387 Til the Cows Come Home (dijkstra）

链接：http://poj.org/problem?id=2387

Til the Cows Come Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46509		Accepted: 15814

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N

* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

Sample Output

90

Hint

INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:

Bessie can get home by following trails 4, 3, 2, and 1.

题意：1 -- n个点，求1--t的最短距离

dijkstra，单源最短路

优化代码：

#define _CRT_SBCURE_MO_DEPRECATE    
#include    
#include    
#include    
#include    
#include    
#include    
#include    
#include    
#include    
#include    
#include     
using namespace std;
const int maxn =2000 + 10;
const int INF =0x3f3f3f3f;


struct node {
	int v, len;
	node(int v = 0, int len = 0) :v(v), len(len) {}
	bool operator < (const node &a)const { //距离从小到大排序  
		return len > a.len;
	}
};
vectorG[maxn];
int dis[maxn];
bool vis[maxn];
int n, t;

void init() {
	for (int i = 0; iQ;
	Q.push(node(s, 0));//加入队列并排序  
	dis[s] = 0;
	while (!Q.empty()) {
		node now = Q.top(); //取出当前最小的  
		Q.pop();
		int v = now.v;
		if (vis[v]) continue; //如果标记过了 直接continue  
		vis[v] = true;
		for (int i = 0; i dis[v] + len) {
				dis[v2] = dis[v] + len;
				Q.push(node(v2, dis[v2]));
			}
		}
	}
	printf("%d\n", dis[e]);
}

int main()
{
	while (scanf("%d %d", &n, &t) != EOF) {
		init();
		for (int i = 1; i <= n; i++) {
			int a, b, len;
			scanf("%d %d %d", &a, &b, &len);
			G[a].push_back(node(b, len));
			G[b].push_back(node(a, len));
		}
		dijkstra_heap(1, t);
	}
	system("pause");
	return 0;
}

为优化版：

错了很多次，因为n和t弄反了，，，用链表不需要判重边

#define _CRT_SBCURE_MO_DEPRECATE    
#include    
#include    
#include    
#include    
#include    
#include    
#include    
#include    
#include    
#include    
#include     
using namespace std;
const int maxn = 2000 + 10;
const int INF = 0x3f3f3f3f;

struct node {
	int vex;
	int weight;
	node(int a, int b) :vex(a), weight(b) {}
};
int n, t;
int mindist[maxn];
bool intree[maxn];
vectorG[maxn];
int m[maxn][maxn];

void init() {
	for (int i = 0; i < maxn; i++) {
		intree[i] = false;
		G[i].clear();
		mindist[i] = INF;
	}
}

void Dijkstra(int s) {
	int ans = 0;
	int vex, addNode, tempMin;
	intree[s] = true;
	mindist[s] = 0;
	for (int i = 0; i < G[s].size(); i++) {
		vex = G[s][i].vex;
		if (mindist[vex] > G[s][i].weight)
			mindist[vex] = G[s][i].weight;
	}
	for (int nodeNum = 1; nodeNum <= n; nodeNum++) {
		tempMin = INF;
		for (int i = 1; i <= n; i++) {
			if (intree[i] == false && mindist[i] < tempMin) {
				tempMin = mindist[i];
				addNode = i;
			}
		}
		intree[addNode] = true;
		for (int i = 0; i < G[addNode].size(); i++) {
			vex = G[addNode][i].vex;
			if (intree[vex] == false && mindist[addNode] + G[addNode][i].weight < mindist[vex])
				mindist[vex] = mindist[addNode] + G[addNode][i].weight;
		}
	}
	printf("%d\n", mindist[n]);
}
int main()
{
	while (scanf("%d %d", &t, &n) != EOF) {
		init();
		for (int i = 1; i <= t; i++) {
			int a, b, len;
			scanf("%d %d %d", &a, &b, &len);
			G[a].push_back(node(b, len));
			G[b].push_back(node(a, len));
		}
		Dijkstra(1);
	}

	return 0;
}