POJ 3268
Silver Cow Party
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 9341 | Accepted: 4210 |
Description
One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?
Input
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Output
Sample Input
4 8 2 1 2 4 1 3 2 1 4 7 2 1 1 2 3 5 3 1 2 3 4 4 4 2 3
Sample Output
10
Hint
解答：
很明显，就是用Dijkstra算法，对G和G的补图进行计算。
第一种办法：代码如下： 该段代码使用邻接矩阵存储。主要是考虑到M可能等于N^2，所以用邻接矩阵可能更省空间。
#include
#include
using namespace std;
//之后可以考虑用临界表做一次
const int N = 1001;
const int MAXTIME = 101*N;//此处是最短路
int Dijkstra(int weight[N][N] ,int n, int sourcever, int result[N]);
int DijkstraT(int weight[N][N] ,int n, int sourcever, int result[N]);
int main() {
int weight[N][N];
memset(weight, 0, sizeof(int) * N * N);
int n, m, sourcever;
cin>>n>>m>>sourcever;
int u, v;
for(int i = 0; i < m; i++) {
cin>>u>>v;
cin>>weight[u][v];
}
int total1[N]; memset(total1, 0, (n + 1) *sizeof(int));
Dijkstra(weight, n, sourcever, total1);
int total2[N]; memset(total2, 0, (n + 1) *sizeof(int));
DijkstraT(weight, n, sourcever, total2);
int max = 0;
for(int i = 1; i <= n; i++) total2[i] += total1[i];
for(int i = 1; i <= n; i++) if(total2[i] > max) max = total2[i];
cout< 代码的效率： 4632k， 268MS
现在改用邻接表做一次：
#include
#include
#include
using namespace std;
const int N = 1001;
const int MAXTIME = 100 * N;
class GNode {
public:
int m_ver;
int m_weight;
GNode *m_next;
GNode(int ver,int weight, GNode *next):m_ver(ver),m_weight(weight), m_next(next){}
};
class Graph {
private:
GNode * m_a[N];
int m_size;
int m_DijResult[N];
public:
Graph(int n);
~Graph();
void m_InsertEdge(int u, int v, int weight);
int *m_Dijkstra(int source);
};
Graph::Graph(int n) {
m_size = n;
memset(m_a, 0, sizeof(GNode *) * (n + 1));
}
Graph::~Graph() {
for(int i = 1; i <= m_size; i++) {
GNode *iter = m_a[i];
while(m_a[i] != 0) {
m_a[i] = iter->m_next;
delete iter;
iter = m_a[i];
}
}
}
void Graph::m_InsertEdge(int u, int v, int weight) {
assert(u <= m_size && u > 0);
assert(v <= m_size && v > 0);
m_a[u] = new GNode(v, weight, m_a[u]);
}
int* Graph::m_Dijkstra(int source) {
bool exist[N];
for(int i = 1; i <= m_size; i++) {
exist[i] = false;
m_DijResult[i] = MAXTIME;
}
m_DijResult[source] = 0;
int count = 0;
while(count < m_size) {
int mintime = MAXTIME, mini;
for(int i = 1; i <= m_size; i++) {
if(!exist[i] && m_DijResult[i] < mintime) {
mintime = m_DijResult[i];
mini = i;
}
}
GNode *iter = m_a[mini];
count++;
exist[mini] = true;
while(iter != 0) {
if(m_DijResult[iter->m_ver] > m_DijResult[mini] + iter->m_weight)
m_DijResult[iter->m_ver] = m_DijResult[mini] + iter->m_weight;
iter = iter->m_next;
}
}
return m_DijResult;
}
int main() {
int n,m,source;
int u, v, weight;
cin>>n>>m>>source;
Graph g1(n), g2(n);
for(int i = 0; i < m; i++) {
cin>>u>>v>>weight;
g1.m_InsertEdge(u, v, weight);
g2.m_InsertEdge(v, u, weight);
}
int *a = g1.m_Dijkstra(source);
int *b = g2.m_Dijkstra(source);
int result[N];
for(int i = 1; i <= n; i++)
result[i] = a[i] + b[i];
int max = 0;
for(int i = 1; i <= n; i++)
if(result[i] > max) max = result[i];
cout< 效率：1260K， 344MS
由此可以推之，在OJ上的测试数据没有完全图。
题目很简单，只是运行了两次Dijkstra算法，但是效率上达不到0MS，是个缺憾。 现在先从基础训练起。