hdu 4009 && POJ 3164

做hdu4009学习了最小树形图 。关于最小树形图的资料可以参看：最小树形图

Transfer water

#include <cstdio>
#include <cstring>
const int maxn = 1005;
struct Edge{
    int s, t, w;
}edge[maxn*maxn];
int ent;//记录边的数量
int n, X, Y, Z;
struct Point{
    int x, y, z;
}pnt[maxn];

void addedge(int s, int t, int w){
    edge[ent].s = s;
    edge[ent].t = t;
    edge[ent].w = w;
    ent++;
}

int absv(int t){
    if(t<0)
        t = -t;
    return t;
}

int manhattan(int i, int j){
    int t = 0;
    t = absv(pnt[i].x - pnt[j].x) +
            absv(pnt[i].y - pnt[j].y)+
            absv(pnt[i].z - pnt[j].z);
    return t;
}

bool read(){
    scanf("%d%d%d%d", &n, &X, &Y, &Z);
    if(!n&&!X&&!Y&&!Z)
        return false;

    for(int i = 1; i <= n; i ++){
        scanf("%d%d%d", &pnt[i].x, &pnt[i].y, &pnt[i].z);
    }

    ent = 0;
    for(int i = 1; i <= n; i ++){
        addedge(0, i, pnt[i].z * X);
    }
    int k, t;
    for(int i = 1; i <= n; i ++){
        scanf("%d", &k);
        while(k--){
            scanf("%d", &t);
            if(pnt[i].z >= pnt[t].z)
                addedge(i, t, manhattan(i,t)*Y);
            else
                addedge(i, t, manhattan(i,t)*Y+Z);
        }
    }
    return true;
}
int in[maxn], pre[maxn], id[maxn], vis[maxn];

int dirmst(int r, int vs){
    int ret = 0;
    while(1){
        int size = vs * 4;
        memset(in, 0x3f, sizeof(in));
        memset(id, -1, sizeof(id));
        memset(vis, -1, sizeof(vis));
        //step1: 找最小入弧
        for(int i = 0; i < ent; i ++){
            int s = edge[i].s;
            int t = edge[i].t;
            int w = edge[i].w;

            if(w >= in[t] || s==t){
                continue;
            }
            pre[t] = s;
            in[t] = w;
        }
        in[r] = 0;
        pre[r] = r;
        for(int i = 0; i < vs ; i ++){
            ret += in[i];
            if(in[i]==0x3f3f3f3f)
                return -1;
        }
        int idx = 0;
        for(int i = 0; i < vs; i ++){
            if(vis[i]==-1){
                int u = i;
                while(vis[u]==-1){
                    vis[u] = i;
                    u = pre[u];
                }
                if(vis[u] != i || u==r)
                    continue;
                for(int t = pre[u]; t != u; t = pre[t])
                    id[t] = idx;
                id[u] = idx ++;
            }
        }
        if(idx==0)
            break;
        for(int i = 0; i < vs; i ++)
            if(id[i]==-1)
                id[i] = idx ++;
        for(int i = 0; i < ent; i ++){
            edge[i].w -= in[edge[i].t];
            edge[i].s  = id[edge[i].s];
            edge[i].t  = id[edge[i].t];
        }
        vs = idx;
        r = id[r];
    }
    return ret;
}

void solve(){
    int ans = dirmst(0, n+1);
    if(ans==-1)
        printf("poor XiaoA\n");
    else
        printf("%d\n", ans);
}

int main(){
    while(read()){
        solve();
    }
    return 0;
}

Command Network

#include <cstdio>
#include <cstring>
#include <cmath>

const int maxn = 105;

struct Point{
    int x, y;
}pnt[maxn];

struct Edge{
    int s, t;
    double w;
}edge[maxn*maxn];
int n, m, ent;

double dist(int i, int j){
    double t1 = 1.0*(pnt[i].x - pnt[j].x);
    double t2 = 1.0*(pnt[i].y - pnt[j].y);
    return sqrt(t1*t1 + t2*t2);
}

void addedge(int s, int t, double d){
    edge[ent].s = s;
    edge[ent].t = t;
    edge[ent].w = d;
    ent ++;
}

bool read(){
    if(scanf("%d%d", &n, &m)==EOF){
        return false;
    }
    for(int i = 1; i <= n; i ++){
        scanf("%d%d", &pnt[i].x, &pnt[i].y);
    }
    ent = 0;
    int s, t;
    for(int i = 1; i <= m; i ++){
        scanf("%d%d", &s, &t);
        addedge(s, t, dist(s,t));
    }
    return true;
}
#define INF 1000000
double in[maxn];
int pre[maxn], id[maxn], vis[maxn];
double dirmst(int r, int vs){
    double ret = 0;
    while(1){
        for(int i = 0; i <= vs; i ++){
            in[i] = INF;
        }
        memset(id, -1, sizeof(id));
        memset(vis, -1, sizeof(vis));

        for(int i = 0; i < ent; i ++){
            int s = edge[i].s;
            int t = edge[i].t;
            double w = edge[i].w;
            if(w>=in[t] || s==t){
                continue;
            }
            pre[t] = s;
            in[t] = w;
        }
        in[r] = 0;
        pre[r] = r;
        for(int i = 1; i < vs; i ++){
            ret += in[i];
            if(in[i]>=INF)
                return -1;
        }
        int idx = 1;
        for(int i = 1; i < vs; i ++){
            if(vis[i] == -1){
                int u = i;
                while(vis[u] == -1){
                    vis[u] = i;
                    u = pre[u];
                }
                if(vis[u] != i || u==r)
                    continue;
                for(int t = pre[u]; t != u; t = pre[t]){
                    id[t] = idx;
                }
                id[u] = idx ++;
            }
        }
        if(idx==1)
            break;
        for(int i = 1; i < vs; i ++){
            if(id[i]==-1){
                id[i] = idx ++;
            }
        }
        for(int i = 0; i < ent; i ++){
            edge[i].w -= in[edge[i].t];
            edge[i].s  = id[edge[i].s];
            edge[i].t  = id[edge[i].t];
        }
        vs = idx;
        r = id[r];
    }
    return ret;
}

void solve(){
    double ans = dirmst(1, n+1);
    if(ans < 0)
        puts("poor snoopy");
    else
        printf("%.2f\n", ans);
}

int main(){
    while(read()){
        solve();
    }
    return 0;
}