Bomb---hdu5934(连通图 缩点)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5934

题意:有n个炸弹,每个炸弹放在(x, y)这个位置,它能炸的范围是以 r 为半径的圆,手动引爆这颗炸弹所需代价是c,当一个炸弹爆炸时,

在它爆炸范围内的所有炸弹都将被它引爆,让求把所有的炸弹都引爆,所需的最少代价是多少?

建立单向图,然后缩点,每个点的权值都为它所在联通块的权值的小的那个,然后找到所有入度为0的点,将他们的权值加起来就是结果;

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
using namespace std;
#define met(a, b) memset(a, b, sizeof(a))
typedef long long LL;
const int N = 1100;
const int INF = 0x3f3f3f3f;
const double eps = 1e-10;

struct node
{
    LL x, y, r;
}a[N];

int n, w[N], Min[N], dfn[N], low[N], vis[N];
int Block[N], nBlock, Stack[N], Top, Time, degree[N];
vector<int> G[N];

void Init()
{
    for(int i=0; i<=n; i++)
        G[i].clear();
    met(Min, INF);///Min[i]表示缩点之后的每个联通块的最小代价;
    met(degree, 0);///记录缩点之后的入度;
    met(dfn, 0);
    met(low, 0);
    met(Stack, 0);
    met(vis, 0);
    met(Block, 0);
    nBlock = Top = Time = 0;
}

void Tajar(int u)
{
    low[u] = dfn[u] = ++Time;
    Stack[Top++] = u;
    vis[u] = 1;
    int v;
    for(int i=0, len=G[u].size(); i<len; i++)
    {
        v = G[u][i];
        if(!dfn[v])
        {
            Tajar(v);
            low[u] = min(low[u], low[v]);
        }
        else if(vis[v])
            low[u] = min(low[u], dfn[v]);
    }

    if(low[u] == dfn[u])
    {
        ++ nBlock;
        do
        {
            v = Stack[--Top];
            Block[v] = nBlock;
            vis[v] = 0;
        }while(u!=v);
    }
}


int main()
{
    int T, t = 1;
    scanf("%d", &T);
    while(T --)
    {
        scanf("%d", &n);

        Init();

        for(int i=1; i<=n; i++)
            scanf("%I64d %I64d %I64d %d", &a[i].x, &a[i].y, &a[i].r, &w[i]);

        for(int i=1; i<=n; i++)///建图
        {
            for(int j=1; j<=n; j++)
            {
                LL d = (a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y);
                if(a[i].r*a[i].r >= d)
                    G[i].push_back(j);
            }
        }

        for(int i=1; i<=n; i++)///缩点;
        {
            if(!dfn[i])
                Tajar(i);
        }

        for(int i=1; i<=n; i++)
        {
            for(int j=0,len=G[i].size(); j<len; j++)
            {
                int x = G[i][j];
                int u = Block[i], v = Block[x];
                if(u != v) degree[v] ++;
                Min[u] = min(Min[u], w[i]);
                Min[v] = min(Min[v], w[x]);
            }
        }

        int ans = 0;
        for(int i=1; i<=nBlock; i++)
        {
            if(degree[i] == 0)
                ans += Min[i];
        }
        printf("Case #%d: %d\n", t++, ans);
    }
    return 0;
}
View Code

 

posted @ 2016-10-29 17:43  西瓜不懂柠檬的酸  Views(284)  Comments(0Edit  收藏  举报
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