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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;

 

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#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;}
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Bomb---hdu5934(连通图 缩点)