hdu5521 Meeting

最短路

我的构图方式是：

将每个集合也抽象成一个点，将集合中的每个点向集合连一条权值为t的有向边，集合向其中的每个点连一条权值为0的有向边，这样 以1和n为起点各做一遍最短路。

枚举中间点便可以得到最后答案了。

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

struct Edge
{
    int v;
    long long t;
};

const int MAXN = 100005;
const int MAXM = 1000005;
const long long INF = 1000000000000010LL;
const int MAXP = MAXM + MAXN;
int n, m, cnt;
int pre[MAXM*2], hd[MAXP];
bool inq[MAXP];
long long ds[MAXP], dt[MAXP];
Edge e[MAXM*2];
vector<int> ansV;

void addEdge(int u, int v, long long t)
{
    pre[cnt] = hd[u];
    hd[u] = cnt;
    e[cnt].v = v;
    e[cnt].t = t;
    cnt++;
}

int main()
{
    int T; scanf("%d", &T);
    for (int kk = 1; kk <= T; kk++)
    {
        //input and initialize
        cnt = 0;
        scanf("%d%d", &n, &m);
        memset(hd, -1, sizeof(hd));
        for (int i = 1; i <= m; i++)
        {
            long long tmp;
            int num;
            scanf("%lld%d", &tmp, &num);
            for (int j = 0; j < num; j++)
            {
                int v; scanf("%d", &v);
                addEdge(v, n + i, tmp);
                addEdge(n + i, v, 0LL);
            }
        }
        
        //spfa-s
        memset(inq, false, sizeof(inq));
        fill(ds, ds + MAXP, INF);
        queue<int> q;
        q.push(1); inq[1] = true; ds[1] = 0LL;
        while (!q.empty())
        {
            int u = q.front(); q.pop();
            int now = hd[u];
            while (now != -1)
            {
                const Edge &ee = e[now];
                if (ds[u] + ee.t < ds[ee.v])
                {
                    ds[ee.v] = ds[u] + ee.t;
                    if (!inq[ee.v])
                    {
                        inq[ee.v] = true;
                        q.push(ee.v);
                    }
                }
                now = pre[now];
            }
            inq[u] = false;
        }
        
        //spfa-t
        memset(inq, false, sizeof(inq));
        fill(dt, dt + MAXP, INF);
        while (!q.empty()) q.pop();
        q.push(n); inq[n] = true; dt[n] = 0LL;
        while (!q.empty())
        {
            int u = q.front(); q.pop();
            int now = hd[u];
            while (now != -1)
            {
                const Edge &ee = e[now];
                if (dt[u] + ee.t < dt[ee.v])
                {
                    dt[ee.v] = dt[u] + ee.t;
                    if (!inq[ee.v])
                    {
                        inq[ee.v] = true;
                        q.push(ee.v);
                    }
                }
                now = pre[now];
            }
            inq[u] = false;
        }
        
        //find the ans & output
        long long ans = INF;
        for (int i = 1; i <= n; i++)
        {
            if (max(ds[i], dt[i]) < ans) ans = max(ds[i], dt[i]);
        }
        ansV.clear();
        if (ans != INF)
        for (int i = 1; i <= n; i++)
        {
            if (max(ds[i], dt[i]) == ans) ansV.push_back(i);
        }
        printf("Case #%d: ", kk);
        if (ans == INF) printf("Evil John\n");
        else
        {
            printf("%lld\n", ans);
            for (int i = 0; i < ansV.size(); i++)
            {
                if (i) printf(" ");
                printf("%d", ansV[i]);
            }
            printf("\n");
        }
    }
    return 0;
}