Invitation Cards POJ 1511 SPFA || dij + heap

http://poj.org/problem?id=1511

求解从1去其他顶点的最短距离之和。

加上其他顶点到1的最短距离之和。

边是单向的。

第一种很容易，直接一个最短路，

然后第二个，需要把边反向建一次，跑一个最短路就好。

★、cin  cout 超时

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
struct HeapNode {
    int u, dis;
    HeapNode(int from, int cost) : u(from), dis(cost) {}
    bool operator < (const struct HeapNode & rhs) const {
        return dis > rhs.dis;
    }
};
const int maxn = 1000000 + 20;
int book[maxn], DFN, dis[maxn];
int first[maxn], num;
struct Node {
    int u, v, w, tonext;
}e[maxn * 2];
void add(int u, int v, int w) {
    ++num;
    e[num].u = u, e[num].v = v, e[num].w = w;
    e[num].tonext = first[u];
    first[u] = num;
}
void dij(int bx, int n) {
    ++DFN;
    for (int i = 1; i <= n; ++i) dis[i] = inf;
    dis[bx] = 0;
    priority_queue<HeapNode> que;
    while (!que.empty()) que.pop();
    que.push(HeapNode(bx, dis[bx]));
    while (!que.empty()) {
        HeapNode t = que.top();
        que.pop();
        int u = t.u;
        if (book[u] == DFN) continue;
        book[u] = DFN;
        for (int i = first[u]; i; i = e[i].tonext) {
            int v = e[i].v;
            if (book[v] != DFN && dis[v] > dis[u] + e[i].w) {
                dis[v] = dis[u] + e[i].w;
                que.push(HeapNode(v, dis[v]));
            }
        }
    }
}
bool in[maxn], tim[maxn];
bool spfa(int bx, int n) { //从bx开始，有n个顶点
    for (int i = 1; i <= n; ++i) {
        dis[i] = inf;
        tim[i] = 0; //入队次数清0
        in[i] = false; //当前这个节点不在队列里
    }
    queue<int> que;
    while (!que.empty()) que.pop();
    que.push(bx), in[bx] = true, dis[bx] = 0, tim[bx]++;
    while (!que.empty()) {
        int u = que.front();
        if (tim[u] > n) return true; //出现负环
        que.pop();
        for (int i = first[u]; i; i = e[i].tonext) {
            if (dis[e[i].v] > dis[e[i].u] + e[i].w) {
                dis[e[i].v] = dis[e[i].u] + e[i].w;
                if (!in[e[i].v]) { //不在队列
                    que.push(e[i].v);
                    in[e[i].v] = true;
                    tim[e[i].v]++;
                }
            }
        }
        in[u] = false;
    }
    return false;
}
int u[maxn], v[maxn], w[maxn];
void init(int n) {
    for (int i = 1; i <= n; ++i) first[i] = 0;
    num = 0;
}
void work() {
    int n, m;
    scanf("%d%d", &n, &m);
    init(n);
    for (int i = 1; i <= m; ++i) {
//        cin >> u[i] >> v[i] >> w[i];
        scanf("%d%d%d", &u[i], &v[i], &w[i]);
        add(u[i], v[i], w[i]);
    }
    spfa(1, n);
    LL ans = 0;
    for (int i = 2; i <= n; ++i) ans += dis[i];
    init(n);
    for (int i = 1; i <= m; ++i) {
        add(v[i], u[i], w[i]);
    }
    spfa(1, n);
    for (int i = 2; i <= n; ++i) ans += dis[i];
    printf("%lld\n", ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}