http://codility.com/demo/take-sample-test/hydrogenium2013
用Dijkstra求最短路径，同时和D[i]比较判断是不是能到。用了优先队列优化，复杂度是(m+n)*log(n)。同时，写Dijkstra的时候一般要用dist数组，这里只拿它做访问标示。中间有个坑就是两个点之间可以多条路径，fail了半天。
#include <queue>
#include <functional> #define pp pair<int,int>
int solution(const vector<int> &A, const vector<int> &B, const vector<int> &C, const vector<int> &D) {
// write your code in C++98
int N = A.size();
int M = D.size();
vector<vector<int> > graph;
graph.resize(M);
for (int i = 0; i < M; i++) {
graph[i].resize(M, -1);
}
for (int i = 0; i < N; i++) {
graph[A[i]][B[i]] = (graph[A[i]][B[i]] == -1 ? C[i] : min(graph[A[i]][B[i]], C[i]));
graph[B[i]][A[i]] = (graph[B[i]][A[i]] == -1 ? C[i] : min(graph[B[i]][A[i]], C[i]));
} vector<int> dist(M, -1);
priority_queue<pp, vector<pp>, greater<pp> > que;
que.push(make_pair(0, 0));
while (!que.empty()) {
pp p = que.top();
que.pop();
if (dist[p.second] == -1) {
dist[p.second] = p.first;
}
else {
continue;
}
if (p.first <= D[p.second]) return p.first;
for (int i = 0; i < graph[p.second].size(); i++) {
if (graph[p.second][i] != -1) {
que.push(make_pair(graph[p.second][i] + p.first, i));
}
}
}
return -1;
}