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Bomb

2024-08-18 03:45:34   223人阅读

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description
There are N bombs needing exploding.

Each bomb has three attributes: exploding radius ri, position (xi,yi) and lighting-cost ci which means you need to pay ci cost making it explode.

If a un-lighting bomb is in or on the border the exploding area of another exploding one, the un-lighting bomb also will explode.

Now you know the attributes of all bombs, please use the minimum cost to explode all bombs.
 
Input
First line contains an integer T, which indicates the number of test cases.

Every test case begins with an integers N, which indicates the numbers of bombs.

In the following N lines, the ith line contains four intergers xi, yi, ri and ci, indicating the coordinate of ith bomb is (xi,yi), exploding radius is ri and lighting-cost is ci.

Limits
- 1≤T≤20
- 1≤N≤1000
- −108≤xi,yi,ri≤108
- 1≤ci≤104
 
Output
For every test case, you should output ‘Case #x: y‘, where x indicates the case number and counts from 1 and y is the minimum cost.
 
Sample Input
150 0 1 51 1 1 60 1 1 73 0 2 105 0 1 4
 
Sample Output
Case #1: 15
分析：对入度为0的强连通分量求一个最小花费；
代码：
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <unordered_map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 1000000007#define inf 0x3f3f3f3f#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, ls[rt]#define Rson mid+1, R, rs[rt]#define sys system("pause")#define freopen freopen("in.txt","r",stdin)const int maxn=1e3+10;using namespace std;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}inline ll read(){    ll x=0;int f=1;char ch=getchar();    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}    return x*f;}int n,m,k,t,pre[maxn],link[maxn],sccno[maxn],du[maxn],dfs_clock,scc_cnt,cas;vi e[maxn],bel[maxn];stack<int>s;struct node{    ll x,y,r,c;    bool operator<(const node&p)const    {        return c<p.c;    }}a[maxn];ll dist(int u,int v){    return (a[u].x-a[v].x)*(a[u].x-a[v].x)+(a[u].y-a[v].y)*(a[u].y-a[v].y);}void dfs(int u){    pre[u]=link[u]=++dfs_clock;    s.push(u);    for(int x:e[u])    {        if(!pre[x])        {            dfs(x);            link[u]=min(link[u],link[x]);        }        else if(!sccno[x])        {            link[u]=min(link[u],pre[x]);        }    }    if(link[u]==pre[u])    {        scc_cnt++;        while(true)        {            int x=s.top();            s.pop();            sccno[x]=scc_cnt;            if(x==u)break;        }    }}void find_scc(int n){    dfs_clock=scc_cnt=0;    memset(sccno,0,sizeof(sccno));    memset(pre,0,sizeof(pre));    for(int i=1;i<=n;i++)        if(!pre[i])dfs(i);}int main(){    int i,j;    scanf("%d",&t);    while(t--)    {        memset(du,0,sizeof(du));        scanf("%d",&n);        rep(i,1,n)bel[i].clear(),e[i].clear();        rep(i,1,n)scanf("%lld%lld%lld%lld",&a[i].x,&a[i].y,&a[i].r,&a[i].c);        rep(i,1,n)rep(j,1,n)        {            if(i==j)continue;            if(dist(i,j)<=a[i].r*a[i].r)e[i].pb(j);        }        find_scc(n);        rep(i,1,n)bel[sccno[i]].pb(i);        rep(i,1,n)rep(j,1,n)        {            if(i==j)continue;            if(dist(i,j)<=a[i].r*a[i].r&&sccno[i]!=sccno[j])++du[sccno[j]];        }        ll ans=0;        rep(i,1,scc_cnt)        {            if(!du[i])            {                ll now=1e18;                for(int x:bel[i])now=min(now,a[x].c);                ans+=now;            }        }        printf("Case #%d: %lld\n",++cas,ans);    }    //system("Pause");    return 0;}

Bomb


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