HDU 4009 transfer water (minimum tree structure: Minimum Spanning Tree template of Directed Graph)

Question:

Link: Click to open the link

Question:

There are n villages that require the minimum cost of water for each village. Each village can obtain water by digging wells or repairing water roads in other villages. The cost of digging a well is multiplied by the height of the House X. The cost of repairing a water track is related to the start point and end point of the directed graph edge.

Ideas:

Code:

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;#define inf 0x3f3f3f3fconst int maxm=1200000;const int maxn=1200;struct point{    int x,y,z;} p[maxn];struct edge{    int u,v,w;} e[maxm];int pre[maxn],id[maxn],vis[maxn],in[maxn];int n,m;void add(int u,int v,int w){    e[m].u=u,e[m].v=v,e[m++].w=w;}int dist(point a,point b){    return abs(a.x-b.x)+abs(a.y-b.y)+abs(a.z-b.z);}int directedMST(int root){    int res=0,nv=n,i;    while (1)    {        for (i=0; i<nv; i++)        {            in[i]=inf;        }        for (i=0; i<m; i++)        {            int u=e[i].u;            int v=e[i].v;            if (e[i].w<in[v]&&u!=v)            {                pre[v]=u;                in[v]=e[i].w;            }        }        for (i=0; i<nv; i++)        {            if (i==root)continue;            if (in[i]==inf)return -1;        }        int cntnode=0;        memset(id,-1,sizeof(id));        memset(vis,-1,sizeof(vis));        in[root]=0;        for (i=0; i<nv; i++)        {            res+=in[i];            int v=i;            while (vis[v]!=i&&id[v]==-1&&v!=root)            {                vis[v]=i;                v=pre[v];            }            if (v!=root&&id[v]==-1)            {                for (int u=pre[v]; u!=v; u=pre[u])                {                    id[u]=cntnode;                }                id[v]=cntnode++;            }        }        if (cntnode==0)            break;        for (i=0; i<nv; i++)        {            if (id[i]==-1)id[i]=cntnode++;        }        for (i=0; i<m; i++)        {            int v=e[i].v;            e[i].u=id[e[i].u];            e[i].v=id[e[i].v];            if (e[i].u!=e[i].v)            {                e[i].w-=in[v];            }        }        nv=cntnode;        root=id[root];    }    return res;}int main(void){    //freopen("input.txt","r",stdin);    int x,y,z;    while(scanf("%d%d%d%d",&n,&x,&y,&z)==4 && (n+x+y+z))    {        m=0;        n++;        for(int i=1; i<n; i++)        {            scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z);            add(0,i,x*p[i].z);        }        for(int i=1; i<n; i++)        {            int k,t;            scanf("%d",&k);            while(k--)            {                scanf("%d",&t);                if(t == i)                    continue;                int cost = dist(p[i],p[t])*y;                if(p[t].z > p[i].z)                    cost += z;                add(i,t,cost);            }        }        printf("%d\n",directedMST(0));    }    return 0;}

-------------------------------------------------------

Fighting, never shrinking; fighting, never stopping ~~~~~~~