HDU5521（最短路+建边）

来源：互联网发布：网络模块接法图解 b 编辑：程序博客网时间：2024/05/25 18:10

图论太菜了啊= =。
题解：每个集合外多加一个点，使得集合中的点和这集合外的点连起来，权值为cost，反过来建一条权值为0的边，然后从1跑spfa，从n跑spfa，那最短路就是max(dist1[i],dist2[i])中的最小值。枚举一下就可以

其实挺水的

#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cstdio>#include <cmath>#include <map>#include <set>#include <queue>#include <vector>#define mod  1000000007#define INF  0x3f3f3f3f#define fuck() (cout << "----------------------------------------" << endl)using namespace std;const int maxn = 2000000 + 5;int dist1[maxn];int dist2[maxn];int queuenum[maxn];bool vis[maxn];int n,m;struct Edge{    int to;    int cost;    Edge(){}    Edge(int To, int Cost)    {        to = To;        cost = Cost;    }};vector <Edge> edge[maxn];void init(){    for(int i=0; i<maxn; i++)        edge[i].clear();    memset(dist1,INF,sizeof(dist1));    memset(dist2,INF,sizeof(dist2));}void addedge(int from, int to, int cost){    edge[from].push_back(Edge(to,cost));}void spfa1(int source){    memset(vis,false,sizeof(vis));    memset(queuenum,0,sizeof(queuenum));    dist1[source] = 0;    queue < int > q;    q.push(source);    vis[source] = true;    queuenum[source] = 1;    while(!q.empty())    {        int u = q.front();        q.pop();        vis[u] = false;        for(int i=0; i<edge[u].size(); i++)        {            Edge eg = edge[u][i];            int to = eg.to;            int cost = eg.cost;            if(dist1[u] + cost < dist1[to])            {                dist1[to] = dist1[u] + cost;                if(!vis[to])                {                    q.push(to);                    queuenum[to]++;                    if(queuenum[to] >= n+m)                        return;                    vis[to] = true;                }            }        }    }    return;}void spfa2(int source){    memset(vis,false,sizeof(vis));    memset(queuenum,0,sizeof(queuenum));    dist2[source] = 0;    queue < int > q;    q.push(source);    vis[source] = true;    queuenum[source] = 1;    while(!q.empty())    {        int u = q.front();        q.pop();        vis[u] = false;        for(int i=0; i<edge[u].size(); i++)        {            Edge eg = edge[u][i];            int to = eg.to;            int cost = eg.cost;            if(dist2[u] + cost < dist2[to])            {                dist2[to] = dist2[u] + cost;                if(!vis[to])                {                    q.push(to);                    queuenum[to]++;                    if(queuenum[to] >= n+m)                        return;                    vis[to] = true;                }            }        }    }    return;}int main(){    int T, kases = 1;    scanf("%d",&T);    while(T--)    {        init();        scanf("%d%d",&n,&m);        for(int i=1; i<=m; i++)        {            int v,num,pt;            scanf("%d%d",&v,&num);            for(int j=0; j<num; j++)            {                scanf("%d",&pt);                addedge(pt,i+n,v);                addedge(i+n,pt,0);            }        }        spfa1(1);        spfa2(n);        int maxcost = INF;        for(int i=1; i<=n; i++)        {            int c = max(dist1[i],dist2[i]);            if(c < maxcost)                maxcost = c;        }        printf("Case #%d: ",kases++);        if(maxcost == INF)        {            printf("Evil John\n");            continue;        }        printf("%d\n",maxcost);        bool flag = true;        for(int i=1; i<=n; i++)        {            int c = max(dist1[i],dist2[i]);            if(c == maxcost)            {                if(flag) printf("%d",i);                else printf(" %d",i);                flag = false;            }        }        printf("\n");    }    return 0;}

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