ZOJ 3691 Flower（最大流+二分）

标签：zoj 3691 flower   最大流   二分   

Gao and his girlfriend‘s relationship becomes better and better after they flying balloon last week. Thus, Gao wants to play a simple but evil game with his girlfriend at a late night.

The game is simple: there is a three-dimensional space, Gao chooses N points to put flowers. For the point i, it has Fi flowers. And his girlfriend has to move all these flowers topoint 1 altogether. As everyone knows, his girlfriend is not strong like Gao, so she can move flowers from one point to another if and only if the Euclidean distance between two points are smaller or equal to R. In another words, she perhaps has to move flowers to the intermediate point for finally moving all flowers to the point one. In order to stay with his girlfriend as much time as possible, he asks his girlfriend, for the point i, that she can only move out no more than Li flowers (including the points as an intermediate point).

Can you help his poor girlfriend to calculate the minimal R?

Input

There are multiple cases.

For each case, the first line contains an integer N (1 ≤ N ≤ 100), which means there are N points.

For the next N lines, each line contains five integers, Xi, Yi, Zi, Fi and Li. Xi, Yi and Zi are the coordinate of point i (0 ≤ Xi, Yi, Zi ≤ 20000), Fi means there are Fi flowers at the beginning. Li means this point can be moved out no more than Li flowers.

Output

For each test case, it contains one real number indicating the minimal R. The results should be rounded to seven decimal places. If there is no solution for this case, please output -1. Output‘s absolute error less than 1e-6 will be accepted.

Sample Input

2
1 1 1 1 1
2 2 2 2 2

Sample Output

1.7320508

但要符合L[i]的限制，但是Gao的女朋友一次能走的距离是R，如果Gao的女朋友把i号点花转移到j号点上，Gao的女朋友需要

行走的距离是dis(i,j)，现在问：Gao的女朋友要完成这项任务需要的最小R，如果完成不了输出-1。

题解：二分答案，建图跑最大流，判断是否满流。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;

const int MAXN = 410;//点数的最大值
const int MAXM = 80010;//边数的最大值
const int INF = 0x3f3f3f3f;

struct Edge {
    int to,next,cap,flow;
} edge[MAXM]; 
int tol,n,m,sum;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];

struct Point {
    double x,y,z;
    int f,l;
} a[MAXN];

double d[MAXN][MAXN];

void init() {
    tol = 0;
    memset(head,-1,sizeof(head));
}

double dis(Point a,Point b) {
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}

void addedge(int u,int v,int w,int rw = 0) {
    edge[tol].to = v;
    edge[tol].cap = w;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = rw;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}

int Q[MAXN];

void BFS(int start,int end) {
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0] = 1;
    int front = 0, rear = 0;
    dep[end] = 0;
    Q[rear++] = end;
    while(front != rear) {
        int u = Q[front++];
        for(int i = head[u]; i != -1; i = edge[i].next) {
            int v = edge[i].to;
            if(dep[v] != -1)continue;
            Q[rear++] = v;
            dep[v] = dep[u] + 1;
            gap[dep[v]]++;
        }
    }
}

int S[MAXN];

int sap(int start,int end,int N) {
    BFS(start,end);
    memcpy(cur,head,sizeof(head));
    int top = 0;
    int u = start;
    int ans = 0;
    while(dep[start] < N) {
        if(u == end) {
            int Min = INF;
            int inser;
            for(int i = 0; i < top; i++)
                if(Min > edge[S[i]].cap - edge[S[i]].flow) {
                    Min = edge[S[i]].cap - edge[S[i]].flow;
                    inser = i;
                }
            for(int i = 0; i < top; i++) {
                edge[S[i]].flow += Min;
                edge[S[i]^1].flow -= Min;
            }
            ans += Min;
            top = inser;
            u = edge[S[top]^1].to;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next) {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]) {
                flag = true;
                cur[u] = i;
                break;
            }
        }
        if(flag) {
            S[top++] = cur[u];
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if(u != start)u = edge[S[--top]^1].to;
    }
    return ans;
}

bool slove(double R) {
    init();
    int s=0,t=1;
    for(int i=2; i<=n; i++) {
        addedge(i,i+n,a[i].l);
        addedge(s,i,a[i].f);
    }
    for(int i=2; i<=n; i++) {
        for(int j=i+1; j<=n; j++) {
            if(d[i][j]<=R) {
                addedge(i+n,j,INF);
                addedge(j+n,i,INF);
            }
        }
        if(d[i][1]<=R)addedge(i+n,1,INF);
    }
    return sum==sap(s,t,2*n);
}

int main() {
    //freopen("test.in","r",stdin);
    while(cin>>n) {
        sum=0;
        for(int i=1; i<=n; i++) {
            scanf("%lf%lf%lf%d%d",&a[i].x,&a[i].y,&a[i].z,&a[i].f,&a[i].l);
            sum+=a[i].f;
        }
        sum-=a[1].f;
        double MaxD=0;
        for(int i=1; i<=n; i++) {
            for(int j=i+1; j<=n; j++) {
                d[i][j]=d[j][i]=dis(a[i],a[j]);
                MaxD=max(MaxD,d[i][j]);
            }
        }
        int flag=0;
        double l=0.0,r=MaxD;
        for(int i=0; i<100; i++) {
            double mid=(l+r)/2;
            if(slove(mid)) {
                r=mid;
                flag=1;
            } else  l=mid;
        }
        if(flag)
            printf("%.7f\n",l);
        else
            printf("-1\n");
    }
    return 0;
}

ZOJ 3691 Flower（最大流+二分）

标签：zoj 3691 flower   最大流   二分   

原文地址：http://blog.csdn.net/acm_baihuzi/article/details/47714785