HDU - 4396 More lumber is required （BFS 最短路）

题目大意：有一个需要采集K跟木头，然后到达点T，现在他从S点出发，路上有N个节点，M条边，只要经过1条边，就可以得到10根木头，问需要花费多少时间才能完成任务

解题思路：木头最多只有500根，且每次收集10跟，最多也只需要收集50次就可以，而且节点最多只有5000个，所以直接暴力BFS即可

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define N 5010
#define M 200010

struct Edgs{
    int to, next, dis;
}E[M];

struct Node{
    int pos, time, wood;
    Node(int pos, int time, int wood): pos(pos), time(time), wood(wood) {}
    bool operator < (const Node &b) const {
        return time > b.time;
    }
};

int n, m, cnt, S, T, K, head[N];
int dp[N][60];

void add_edgs(int u, int v, int l) {
    E[cnt].to = v;
    E[cnt].dis = l;
    E[cnt].next = head[u];
    head[u] = cnt++;
}

void init() {
    memset(head, -1, sizeof(head));
    cnt = 0;

    int x, y, z;
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &x, &y, &z);
        add_edgs(x, y, z);
        add_edgs(y, x, z);
    }
}

int bfs() {
    memset(dp, 0x3f, sizeof(dp));
    priority_queue<Node> q;
    q.push(Node(S, 0, 0));
    dp[S][0] = 0;

    while (!q.empty()) {
        Node t = q.top();
        q.pop();
        int u = t.pos, time = t.time, wood = t.wood;

        if (u == T && wood >= K)
            return time;

        for (int i = head[u]; i != -1; i = E[i].next) {
            int v = E[i].to;
            if (dp[v][wood + 1] > dp[u][wood] + E[i].dis) {
                dp[v][wood + 1] = dp[u][wood] + E[i].dis;
                q.push(Node(v, dp[v][wood + 1], wood + 1));
            }
        }
    }
    return -1;
}

void solve() {
    scanf("%d%d%d", &S, &T, &K);
    if (K % 10)
        K = K / 10 + 1;
    else
        K = K / 10;
    printf("%d\n", bfs());
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        init();
        solve();
    }
    return 0;
}

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