HDU 5934 Bomb 【强连通缩点】

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

1
5
0 0 1 5
1 1 1 6
0 1 1 7
3 0 2 10
5 0 1 4

 

Sample Output

Case #1: 15
对图进行强联通后缩点寻找入度为0的点，在入度为0的联通快中寻找价值最小的的点值即为该强连通分量的值；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mms(x, y) memset(x, y, sizeof x)
const int MAX = 1500;
const int INF = 0x3f3f3f3f;
struct node
{
    ll x, y, r;
    int c;
} a[MAX];
struct edge
{
    int to, next;
} e[MAX * MAX * 2];
int num, cnt, pos, sig, n;
bool vis[MAX];
int ans[MAX];
int dfn[MAX];
int low[MAX];
int in[MAX];
int head[MAX];
int col[MAX];
int Stack[MAX];
void add(int u, int v)
{
    e[num] = {v, head[u]};
    head[u] = num++;
}
void init()
{
    mms(ans, INF);
    mms(vis, 0);
    mms(low, 0);
    mms(dfn, 0);
    mms(in, 0);
    mms(col, 0);
    mms(Stack, 0);
    cnt = pos = sig = 0;
}
void tarjan(int u)
{
    int v;
    low[u] = dfn[u] = ++cnt;
    Stack[pos++] = u;
    vis[u] = 1;
    for(int i = head[u]; i != -1; i = e[i].next)
    {
        v = e[i].to;
        if(!dfn[v])
        {
            tarjan(v);
            if(low[u] > low[v])
                low[u] = low[v];
        }
        else if(vis[v] && low[u] > dfn[v])
            low[u] = dfn[v];
    }
    if(low[u] == dfn[u])
    {
        sig++;
        do
        {
            v = Stack[--pos];
            vis[v] = 0;
            col[v] = sig;
        }
        while(u != v);
    }
}
int main()
{
    int N, kase = 0;
    scanf("%d", &N);
    while(N--)
    {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)
            scanf("%lld%lld%lld%lld", &a[i].x, &a[i].y, &a[i].r, &a[i].c);
        num = 0;
        mms(head, -1);
        for(int i = 1; i <= n; i++)
        {
            for(int j = i + 1; j <= n; j++)
            {
                ll x = a[i].x - a[j].x;
                ll y = a[i].y - a[j].y;
                ll dis = x * x + y * y;
                ll dis1 = a[i].r * a[i].r;
                ll dis2 = a[j].r * a[j].r;
                if(dis <= dis1)
                    add(i, j);
                if(dis <= dis2)
                    add(j, i);
            }
        }
        init();
        for(int i = 1; i <= n; i++)
            if(!dfn[i])
                tarjan(i);
        for(int i = 1; i <= n; i++)
        {
            for(int j = head[i]; j != -1; j = e[j].next)
            {
                int x = e[j].to;
                if(col[i] != col[x])
                    in[col[x]]++;
            }
        }
        for(int i = 1; i <= n; i++)
        {
            int x = col[i];
            if(in[x] == 0)
                ans[x] = min(ans[x], a[i].c);
        }
        int res = 0;
        for(int i = 1; i <= sig; i++)
            if(in[i] == 0)
                res += ans[i];
        printf("Case #%d: %d\n", ++kase, res);
    }
    return 0;
}