最短路(Dijkstra) HDOJ 4318 Power transmission

题目传送门

题意：起点s到终点t送电，中途会有损耗，问最小损耗是多少

分析：可以转换为单源最短路问题，用优先队列的Dijkstra版本，d[]表示从s出发到当前点的最小损耗，用res保存剩下的电量。当到达t就结束，因为按照w从小到大排序，访问过的点都已经最优，这是贪心思想

收获：复习了Dijkstra，进一步理解Dijkstra的贪心的思想(程序跑得慢，还可优化)

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-24 9:51:51
* File Name     :I.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
typedef pair<int, int> P;
const int N = 5e4 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Edge	{
	int v;
	double w, res;
	bool operator < (const Edge &r)	const {
		return w > r.w;
	}
};
vector<Edge> G[N];
bool vis[N];
double d[N];
int n, s, t;
double M;

void Dijkstra(void)	{
	priority_queue<Edge> Q;
    memset (vis, false, sizeof (vis));
	for (int i=1; i<=n; ++i)	d[i] = INF;
    d[s] = 0;	Q.push ((Edge) {s, 0, M});
    while (!Q.empty ())	{
		Edge r = Q.top ();	Q.pop ();
		int u = r.v; double res = r.res;
		if (vis[u]) continue;
		vis[u] = true;
		if (u == t)	{
			printf ("%.2f
", M - res);	return ;
		}
		for (int i=0; i<G[u].size (); ++i)	{
			int v = G[u][i].v;	double w = G[u][i].w;
			if (!vis[v] && d[v] > d[u] + res * w)	{
				d[v] = d[u] + res * w;
				Q.push ((Edge) {v, d[v], res - res * w});
			}
		}
    }

    puts ("IMPOSSIBLE!");
}

int main(void)    {
	while (scanf ("%d", &n) == 1)	{
		for (int i=1; i<=n; ++i)	{
			G[i].clear ();
			int k;	scanf ("%d", &k);
			for (int j=1; j<=k; ++j)	{
				int v;	double w;	scanf ("%d%lf", &v, &w);	w /= 100;
				G[i].push_back ((Edge) {v, w});
			}
		}
		scanf ("%d%d%lf", &s, &t, &M);
		Dijkstra ();
	}

    return 0;
}