[hdu 5934 Bomb]Tarjan强连通分量

[hdu 5934 Bomb]Tarjan强连通分量

题目链接：[hdu 5934 Bomb]

题意描述：

有N个炸弹,第i个炸弹的坐标是(xi,yi)， 半径为ri，手动将其引爆的花费是ci。如果炸弹a在另外一个炸弹b的爆炸范围内，那么b的爆炸会引起a的爆炸。求让所有的炸弹爆炸的最小花费。

解题思路：

如果炸弹a在另外一个炸弹b的爆炸范围内，那么就添加一条a到b的有向边，然后对图缩点，统计强连通分量中最小的顶点权值，以及每个缩点的入度。然后答案就是,所有入度为0的缩点中的最小的顶点权值之和。

#include <bits/stdc++.h>
using namespace std;

typedef __int64 LL;
typedef pair<LL, LL> PLL;

const int MAXN = 1000 + 5;

int T, cas, N;
struct QNode {
LL x, y, r, c, id;
} q[MAXN];

vector<int> G[MAXN];

int In[MAXN]; // 入度
int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN]; //Belong数组的值是1~scc
int Index, top;
int scc;//强连通分量的个数
bool Instack[MAXN];
int num[MAXN];//各个强连通分量包含点的个数，数组编号1~scc
LL w[MAXN]; // 联通分量中最小值

void Tarjan(int u) {
int v, sz = G[u].size();
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;

for(int i = 0; i < sz; i++) {
v = G[u][i];
if(!DFN[v]) {
Tarjan(v);
if( Low[u] > Low[v] )Low[u] = Low[v];
} else if(Instack[v] && Low[u] > DFN[v])
Low[u] = DFN[v];
}
if(Low[u] == DFN[u]) {
scc++;
do {
v = Stack[--top];
Instack[v] = false;
Belong[v] = scc;
num[scc]++;
} while( v != u);
}
}

void work() {
memset(DFN, 0, sizeof(DFN));
memset(Instack, false, sizeof(Instack));
memset(num, 0, sizeof(num));
memset(In, 0, sizeof(In));

Index = scc = top = 0;
for(int i = 1; i <= N; i++)
if(!DFN[i])
Tarjan(i);

memset(w, 0x3f, sizeof(w));

for(int u = 1; u <= N; u++) {
int x = Belong[u], y, v, sz = G[u].size();;
w[x] = min(w[x], q[u].c);
for(int j = 0; j < sz; j++) {
v = G[u][j], y = Belong[v];
if(x == y) continue;
In[y] ++;
}
}
}

inline bool IsInter(QNode& a, QNode& b) {
LL dist = (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
return dist <= a.r * a.r;
}

int main() {
scanf("%d", &T); cas = 0;
while(T --) {
scanf("%d", &N);
for(int i = 1; i <= N; i++) {
scanf("%I64d %I64d %I64d %I64d", &q[i].x, &q[i].y, &q[i].r, &q[i].c);
}
for(int i = 1; i <= N; i++) {
G[i].clear();
for(int j = 1; j <= N; j++) {
if (j == i) continue;
if(IsInter(q[i], q[j])) {
G[i].push_back(j);
}
}
}
work();
LL ans = 0;
for(int i = 1; i <= scc; i++) {
if(In[i] == 0) {
ans += w[i];
}
}
printf("Case #%d: %I64d\n", ++cas, ans);
}
return 0;
}