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; }