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  • Countries in War 强连通

      强连通分量之间权值为0

    强连通加最短路

    //#include<bits/stdc++.h>
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    //input by bxd
    #define rep(i,a,b) for(int i=(a);i<=(b);i++)
    #define repp(i,a,b) for(int i=(a);i>=(b);--i)
    #define RI(n) scanf("%d",&(n))
    #define RII(n,m) scanf("%d%d",&n,&m)
    #define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
    #define RS(s) scanf("%s",s);
    #define ll long long
    #define pb push_back
    #define REP(i,N)  for(int i=0;i<(N);i++)
    #define CLR(A,v)  memset(A,v,sizeof A)
    //////////////////////////////////
    #define inf 0x3f3f3f3f
    const int N=500+5;
    const int M=N*N;
    int head[M],pos;
    struct Edge
    {
        int nex,to,v;
    }edge[M];
    struct node
    {
        int s,e,v;
    }s[M];
    void add(int a,int b,int c)
    {
        edge[++pos].nex=head[a];
        head[a]=pos;
        edge[pos].v=c;
        edge[pos].to=b;
    }
    int tot,ind,cnt,Stack[N],dfn[N],low[N],vis[N],belong[N];
    int dis[N];
    int mp[N][N];
    int n,m;
    void init()
    {
        tot=ind=cnt=pos=0;
        CLR(Stack,0);
        CLR(head,0);
        CLR(vis,0);
        CLR(dfn,0);
        CLR(low,0);
        rep(i,1,n)
        rep(j,1,n)
        if(i!=j)mp[i][j]=inf;
        else mp[i][j]=0;
    
    }
    void tarjan(int x)
    {
        dfn[x]=low[x]=++tot;
        Stack[++ind]=x;
        vis[x]=1;
        for(int i=head[x];i;i=edge[i].nex)
        {
            int v=edge[i].to;
            if(!dfn[v])
            {
                tarjan(v);
                low[x]=min(low[x],low[v]);
            }
            else if(vis[v])
                low[x]=min(low[x],low[v]);
        }
        if(dfn[x]==low[x])
        {
            cnt++;int v;
            do
            {
                v=Stack[ind--];
                vis[v]=0;
                belong[v]=cnt;
            }
            while(x!=v);
        }
    }
    
    
    int main()
    {
        while(RII(n,m),n||m)
        {
            init();
            rep(i,1,m)
            {
                int a,b,c;RIII(a,b,c);
    
                add(a,b,c);
            }
            rep(i,1,n)
            if(!dfn[i])tarjan(i);
    
    
            rep(i,1,n)
            {
                int u=belong[i];
                for(int j=head[i];j;j=edge[j].nex)
                {
                    int v=belong[ edge[j].to ];
                    if(u!=v)
                    {
                        mp[u][v]=min(mp[u][v],edge[j].v);
                    }
                }
            }
    
            int k;
            RI(k);
            int last=-1;
            while(k--)
            {
                int a,b;
                RII(a,b);
                a=belong[a];
                b=belong[b];
                if(a==last)
                {
                    if(dis[b]==inf)
                        printf("Nao e possivel entregar a carta
    ");
                    else
                        printf("%d
    ",dis[b]);
                }
                else
                {
                    last=a;
                    rep(i,1,cnt)
                    dis[i]=inf;
                    dis[a]=0;
                    CLR(vis,0);
                    while(1)
                    {
                        int u=-1,minn=inf;
                        rep(i,1,cnt)
                        if(!vis[i]&&dis[i]<minn)
                            minn=dis[u=i];
                        if(u==-1)break;
                        vis[u]=1;
                        rep(i,1,cnt)
                        dis[i]=min(dis[i],dis[u]+mp[u][i]);
                    }
                    if(dis[b]==inf)
                        printf("Nao e possivel entregar a carta
    ");
                    else
                        printf("%d
    ",dis[b]);
                }
            }
            cout<<endl;
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/bxd123/p/10799790.html
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