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最新推荐文章于 2020-07-29 10:10:39 发布

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本文介绍了一种基于图论的算法解决炸弹连锁爆炸问题的方法。通过构建图模型并利用Tarjan算法寻找强连通分量，进而求得引爆所有炸弹所需的最小成本。详细解释了算法流程，包括输入输出格式及示例。

here are NN bombs needing exploding.

Each bomb has three attributes: exploding radius riri, position (xi,yi)(xi,yi) and lighting-cost cici which means you need to pay cici 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 TT, which indicates the number of test cases.

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

In the following NN lines, the ith line contains four intergers xixi, yiyi, riri and cici, indicating the coordinate of ith bomb is (xi,yi)(xi,yi), exploding radius is riri and lighting-cost is cici.

Limits
- 1≤T≤201≤T≤20
- 1≤N≤10001≤N≤1000
- −108≤xi,yi,ri≤108−108≤xi,yi,ri≤108
- 1≤ci≤1041≤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

tarjan以后找根

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int long long 
int liantong[1001], dfn[1001], low[1001], clj = 0, jc = 0;
stack<int>zz;
int inf=0x3f3f3f3f3f3f3f3f;
int T,n,x[1001],y[1001],r[1001],c[1001],u=0,chu[1001],dan[1001];
vector<int>tu[1001];
void tarjan(int xianzai)
{
    dfn[xianzai] = low[xianzai] = ++clj;
    zz.push(xianzai);
    for (int a = 0;a<tu[xianzai].size();a++)
    {
        int dd = tu[xianzai][a];
        if (!dfn[dd])
        {
            tarjan(dd);
            low[xianzai] = min(low[xianzai], low[dd]);
        }
        else if (!liantong[dd])low[xianzai] = min(low[xianzai], low[dd]);
    }
    if (low[xianzai] == dfn[xianzai])
    {
        jc++;
        for (;;)
        {
            int ww = zz.top();
            zz.pop();
            liantong[ww] = jc;
            if (ww == xianzai)break;
        }
    }
}
main()
{
    cin>>T;
    while(T--)
    {
        cin>>n;
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(liantong,0,sizeof(liantong));
        memset(r,0,sizeof(r));
        memset(c,0,sizeof(c));
        memset(chu,0,sizeof(chu));
        memset(dan,0x3f,sizeof(dan));
        clj=jc=0;
        while(!zz.empty())zz.pop();
        for(int a=1;a<=n;a++)tu[a].clear();
        for(int a=1;a<=n;a++)scanf("%lld%lld%lld%lld",&x[a],&y[a],&r[a],&c[a]);
        for(int a=1;a<=n;a++)
        {
            for(int b=1;b<=n;b++)
            {
                if(b==a)continue;
                if((x[b]-x[a])*(x[b]-x[a])+(y[a]-y[b])*(y[a]-y[b])>r[a]*r[a])continue;
                tu[a].push_back(b);
            }
        }
        for(int a=1;a<=n;a++)
        {
            if(liantong[a])continue;
            tarjan(a);
        }
        for(int a=1;a<=n;a++)
        {
            for(int b=0;b<tu[a].size();b++)
            {
                int w=tu[a][b];
                if(liantong[a]==liantong[w])continue;
                chu[liantong[w]]++;
            }
        }
        for(int a=1;a<=n;a++)
        {
            if(chu[liantong[a]])continue;
            dan[liantong[a]]=min(dan[liantong[a]],c[a]);
        //  cout<<liantong[a]<<endl;
        }
        /*for(int a=1;a<=jc;a++)
        {
            cout<<chu[a]<<endl;
        }
        return 0;*/
        int dandan=0;
        for(int a=1;a<=jc;a++)
        {
            if(dan[a]==inf)continue;
            if(chu[a])continue;
            dandan+=dan[a];
        }
        printf("Case #%lld: %lld\n",++u,dandan);
    }
}