Dijkstra算法是求单源最短路径好方法,但是只能处理没有负边的图的问题,有一条边为负的就会导致最终的结果不正确
时间复杂度可以达到O(|E|log|v|), 所以一般求单源最短路径问题都可以用这个算法
伪代码:
priority_queue que;
que.push(startV);
while que is not empty
V a = que.top();
que.pop();
for all V connect with a:
if d(v) > d(a) + length_a_to_v;
d(v) = d(a) + length_a_to_v;
que.push(v);
直接上代码:
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#include <functional>
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#include <queue>
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using namespace std;
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#define MAXV 4
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#define INF 100000
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int d[4] = { 0 };
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int edge[4][4] = { 0 };
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void init()
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{
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for (int i = 0; i < MAXV; i++)
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for (int j = 0; j < MAXV; j++)
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edge[i][j] = INF;
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edge[0][0] = 0;
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edge[0][1] = 6;
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edge[0][2] = 1;
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edge[1][3] = 2;
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edge[2][3] = 8;
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edge[3][0] = 3;
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}
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typedef pair<int, int> p;
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void dijsktra(p s)
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{
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priority_queue<p, vector<p>, greater<p>> que;
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que.push(s);
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while (!que.empty())
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{
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p tmp = que.top();
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que.pop();
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for (int i = 0; i < MAXV; i++)
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{
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if (tmp.first != i)
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{
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if (edge[tmp.first][i] != INF && (d[i] > d[tmp.first] + edge[tmp.first][i]))
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{
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d[i] = d[tmp.first] + edge[tmp.first][i];
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p newP;
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newP.first = i;
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newP.second = d[i];
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que.push(newP);
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}
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}
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}
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}
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for (int i = 0; i < MAXV; i++)
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{
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printf("%d",d[i]);
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}
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}
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int _tmain(int argc, _TCHAR* argv[])
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{
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fill(d, d + 4, INF);
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init();
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d[0] = 0;
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p s;
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s.first = 0;
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s.second = d[0];
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dijsktra(s);
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return 0;
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}
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