Dijkstra: Till the Cows Come Home
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D - Til the Cows Come Home
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uDescription
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.
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.
* 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 51 2 202 3 303 4 204 5 201 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.
堆优化dijkstra
There are five landmarks.
OUTPUT DETAILS:
Bessie can get home by following trails 4, 3, 2, and 1.
#define INF (1<<30)-1int w[2010][2010];int v[2010];int d[2010];int t, n;void init(){ int i, j; memset(v, 0, sizeof(v)); for(i = 1; i <= n; i ++) for(j = 1; j <= n; j ++){ //预处理两点间距离，无限大即没连通 w[i][j] = (i == j ? 0 : INF); } for(i = 1; i <= n; i ++) //预处理出发点到i点的距离 d[i] = (i == 1 ? 0 : INF);}void dijkstra(){ int i, j, x, m; for(i = 1; i <= n; i ++){ //循环n次，里面跟用Prim算法最小生成树类似 m = INF; for(j = 1; j <= n; j ++){ //找到没到过的到起点最近的点 if(!v[j] && d[j] <= m) m = d[x=j]; } v[x] = 1; //标记新加入的点 for(j = 1; j <= n; j ++){ //由于新扩展的点导致某些点到起点的距离更新，而Prim算法是到树的距离的更新 if(d[j] > d[x] + w[x][j]) d[j] = d[x] + w[x][j]; } } cout << d[n] << endl;}int main(){ int i, j, ww, y; int z; while(cin >> t >> n){ init(); for(i = 0; i < t; i ++){ //输入的时候注意有重复的边 scanf("%d %d", &z, &y); if(w[z][y] == INF && w[z][y] == INF){ scanf("%d", &w[z][y]); w[y][z] = w[z][y]; } else{ scanf("%d", &ww); if(ww < min(w[z][y], w[y][z])){ w[z][y] = ww; //因为是无向图所以zy yz距离一样 w[y][z] = ww; } } } dijkstra(); } return 0;}堆优化dijkstra
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <algorithm>using namespace std;#define INF 1 << 31 - 1int t, n;int cnt;int lastshow[40010];int d[40010];typedef pair<int, int> pii;priority_queue<pii, vector<pii>, greater<pii> > q;struct edge{ int to; int wei; int next;}e[40010];void insert(int a, int b, int c){ cnt ++; e[cnt].to = b; e[cnt].next = lastshow[a]; e[cnt].wei = c; lastshow[a] = cnt;}bool done[40010];void dijkstra(){ memset(done, 0, sizeof(done)); q.push(make_pair(d[1], 1)); while(!q.empty()){ pii u = q.top(); q.pop(); int x = u.second; if(done[x]) continue; done[x] = true; for(int i = lastshow[x]; i != -1; i = e[i].next){ if(d[e[i].to] > d[x] + e[i].wei){ d[e[i].to] = d[x] + e[i].wei; q.push(make_pair(d[e[i].to], e[i].to)); } } }}int main(){ int i, j, k, m, ww, x, y; int z; int cnt; while(cin >> t >> n){ cnt = 0; memset(lastshow, -1, sizeof(lastshow)); for(i = 1; i <= n; i ++) //预处理出发点到i点的距离 d[i] = (i == 1 ? 0 : INF); for(i = 0; i < t; i ++){ //输入的时候不用注意有重复的边 int a, b, c; scanf("%d %d %d", &a, &b, &c); insert(a, b, c); insert(b, a, c); } dijkstra(); cout << d[n] << endl; } return 0;}- Dijkstra: Till the Cows Come Home
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