[二分+最大流]zoj 3691：flower

大致题意：

    在一个三维空间中有n个点，每个点的坐标都是正整数。第i个点上面有fi个花朵，每个点最多能运送li朵花到其他的点上，每次只能把一个点上的花朵送到距离其R的点上。现在求出最小的R，使得所有的花朵都能被移动到第一个点上。

 

大致思路：

    二分+最大流~~好久没ac的说

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int inf=1<<28;
const int nMax=800;
const int mMax=100000;

class node{
    public:
    int c,u,v,next;
};node edge[mMax];
int ne, head[nMax];
int cur[nMax], ps[nMax], dep[nMax];

void addedge(int u, int v,int w){   ////dinic
    edge[ne].u = u;
    edge[ne].v = v;
    edge[ne].c = w;
    edge[ne].next = head[u];
    head[u] = ne ++;
    edge[ne].u = v;
    edge[ne].v = u;
    edge[ne].c = 0;
    edge[ne].next = head[v];
    head[v] = ne ++;
}

int dinic(int s, int t){                       //  dinic
    int tr, res = 0;
    int i, j, k, f, r, top;
    while(1){
        memset(dep, -1, sizeof(dep));
        for(f = dep[ps[0]=s] = 0, r = 1; f != r;)
            for(i = ps[f ++], j = head[i]; j; j = edge[j].next)
                if(edge[j].c && dep[k=edge[j].v] == -1){
                    dep[k] = dep[i] + 1;
                    ps[r ++] = k;
                    if(k == t){
                        f = r; break;
                    }
                }
        if(dep[t] == -1) break;
        memcpy(cur, head, sizeof(cur));
        i = s, top = 0;
        while(1){
            if(i == t){
                for(tr =inf, k = 0; k < top; k ++)
                    if(edge[ps[k]].c < tr)
                        tr = edge[ps[f=k]].c;
                for(k = 0; k < top; k ++){
                    edge[ps[k]].c -= tr;
                    edge[ps[k]^1].c += tr;
                }
                i = edge[ps[top=f]].u;
                res += tr;
            }
            for(j = cur[i]; cur[i]; j = cur[i] =edge[cur[i]].next){
                if(edge[j].c && dep[i]+1 == dep[edge[j].v]) break;
            }
            if(cur[i]){
                ps[top ++] = cur[i];
                i = edge[cur[i]].v;
            }
            else{
                if(top == 0) break;
                dep[i] = -1;
                i = edge[ps[-- top]].u;
            }
        }
    }
    return res;
}

//double max(double a,double b){if(a>b)return a;return b;}
int sum[nMax][5],num,n;
int dis[nMax][nMax];
void getdis(int a,int b){
    if(a==b)return;
    int res;
    res=(sum[a][0]-sum[b][0])*(sum[a][0]-sum[b][0])+(sum[a][1]-sum[b][1])*(sum[a][1]-sum[b][1])+(sum[a][2]-sum[b][2])*(sum[a][2]-sum[b][2]);
    dis[a][b]=res;
}

bool check(int mid){
    int i,j,res;
    ne=2;
    memset(head,0,sizeof(head));
    for(i=1;i<n;i++){
        addedge(0,i,sum[i][3]);
        addedge(i,i+n,sum[i][4]);
        for(j=1;j<n;j++){
            if(dis[i][j]<=mid){
                addedge(i+n,j,sum[i][3]);
            }
        }
        if(dis[i][0]<=mid)addedge(i+n,n*2+1,sum[i][4]);
    }
    res=dinic(0,2*n+1);
    if(res>=num)return 1;
    return 0;
}

int main(){
    int i,j,ans;
    while(cin>>n){
        num=0;
        for(i=0;i<n;i++){
            scanf("%d%d%d%d%d",&sum[i][0],&sum[i][1],&sum[i][2],&sum[i][3],&sum[i][4]);
            if(i!=0)num+=sum[i][3];
        }
        if(n==1){
            printf("%.7lf\n",0);
            continue;
        }
        int left=0,right,mid;
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                getdis(i,j);
                right=max(right,dis[i][j]);
            }
        }
        if(!check(right)){
            cout<<-1<<endl;
            continue;
        }
        while(right>=left){
            mid=(left+right)/2;
            if(check(mid)){
                right=mid-1;
                ans=mid;
            }
            else{
                left=mid+1;
            }
        }
        printf("%.7lf\n",sqrt(ans*1.0));
    }
    return 0;
}

 

评论

1 楼 McFlurry 2013-05-02  

YM宇哥，宝刀未老。