POJ-1511 Invitation Cards 最短路

题意：求一个点到所有点的最短距离之和加上所有点到这个点距离之和，边为单向边。

解法：先做一次spfa，然后将所有的边反序做一次spfa即可。

代码如下：

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

const int MaxN = 1000005;

int P, Q, idx, gdis[MaxN], cdis[MaxN];
int head[MaxN], rhead[MaxN];
bool vis[MaxN];

struct Edge {
    int v, fee, next;
}e[MaxN], re[MaxN];

void insert(int a, int b, int c) {
    ++idx;
    e[idx].v = b, e[idx].fee = c;
    e[idx].next = head[a];
    head[a] = idx;
    
    re[idx].v = a, re[idx].fee = c;
    re[idx].next = rhead[b];
    rhead[b] = idx;
}

void spfa(Edge e[], int head[], int dis[]) {
    queue<int>q;
    memset(vis, 0, sizeof (vis));
    dis[1] = 0;
    vis[1] = true;
    q.push(1);
    while (!q.empty()) {
        int v = q.front();
        q.pop();
        vis[v] = false;
        for (int i = head[v]; i != -1; i = e[i].next) {
            if (dis[e[i].v] > dis[v] + e[i].fee) {
                dis[e[i].v] = dis[v] + e[i].fee;
                if (!vis[e[i].v]) {
                    q.push(e[i].v);
                    vis[e[i].v] = true;
                }
            }
        }
    }
}

int main() {
    int T;
    cin >> T;
    while (T--) {
        int a, b, c;
        memset(head, 0xff, sizeof (head));
        memset(rhead, 0xff, sizeof (rhead));
        idx = 0;
        scanf("%d %d", &P, &Q);
        for (int i = 1; i <= P; ++i) {
            gdis[i] = cdis[i] = 0x7fffffff;    
        }
        for (int i = 0; i < Q; ++i) {
            scanf("%d %d %d", &a, &b, &c);
            insert(a, b, c);
        }
        spfa(e, head, gdis);
        spfa(re, rhead, cdis);
        long long tot = 0;
        for (int i = 2; i <= P; ++i) {
            tot += gdis[i] + cdis[i];
        }
        cout << tot << endl;
    }    
    return 0;    
}