poj3159 Candies

题意：

有N个孩子（N<=3000)分糖果。
有M个关系(M<=150,000)。每个关系形如：
A B C
表示第B个学生比第A个学生多分到的糖果数目，不能超过C
求第N个学生最多比第1个学生能多分几个糖果

思路：

dijkstra+heap

实现：

#include <iostream>
#include <vector>
#include <queue>
#include <functional>
#include <cstdio>
using namespace std;

const int INF = 0x3f3f3f3f;

struct edge
{
    int to;
    int cost;
};
vector<edge> G[30005];
int d[30005];
typedef pair<int, int>P;

int Scan()
{
    int res = 0, ch, flag = 0;

    if ((ch = getchar()) == '-')             
        flag = 1;

    else if (ch >= '0' && ch <= '9')          
        res = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9')
        res = res * 10 + ch - '0';

    return flag ? -res : res;
}

int main()
{
    int n, m, a, b, l;
    cin >> n >> m;
    for (int i = 0; i < m; i++)
    {
        a = Scan();
        b = Scan();
        l = Scan();
        edge e;
        e.to = b;
        e.cost = l;
        G[a].push_back(e);
    }
    priority_queue<P, vector<P>, greater<P> >myQueue;
    myQueue.push(P(0, 1));
    for (int i = 1; i <= n; i++)
    {
        d[i] = INF;
    }
    d[1] = 0;
    while (!myQueue.empty())
    {
        P temp = myQueue.top();
        myQueue.pop();
        int v = temp.second;
        if (d[v] < temp.first)
            continue;
        for (int i = 0; i < G[v].size(); i++)
        {
            edge e = G[v][i];
            if (d[e.to] > d[v] + e.cost)
            {
                d[e.to] = d[v] + e.cost;
                myQueue.push(P(d[e.to], e.to));
            }
        }
    }
    cout << d[n] << endl;
    return 0;
}