[HDOJ5934]Bomb（强连通分量，缩点）

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

题意：有n个炸弹，爆炸范围和点燃花费给你，如果一个爆炸那么它爆炸范围内的炸弹也会爆炸。问让所有炸弹爆炸的最小花费。

遍历任意两个炸弹，如果i在j的爆炸范围内，则建一条有向边。缩完点以后找入度为0的点点燃就行了。

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

 typedef long long LL;
 typedef struct Edge {
     int u;
     int v;
     int next;
     Edge() { next = -; }
 }Edge;
 typedef struct P {
     LL x, y, r;
     int c;
 }P;
 const int maxn = ;
 P p[maxn];

 int head[maxn], ecnt;
 Edge edge[maxn*maxn];
 int n, m;

 int bcnt, dindex;
 int dfn[maxn], low[maxn];
 int stk[maxn], top;
 int belong[maxn];
 bool instk[maxn];
 int ret[maxn];
 int in[maxn];

 void init() {
     memset(edge, , sizeof(edge));
     memset(head, -, sizeof(head));
     memset(instk, , sizeof(instk));
     memset(dfn, , sizeof(dfn));
     memset(low, , sizeof(low));
     memset(belong, , sizeof(belong));
     ecnt = top = bcnt = dindex = ;
 }

 void adde(int uu, int vv) {
     edge[ecnt].u = uu;
     edge[ecnt].v = vv;
     edge[ecnt].next = head[uu];
     head[uu] = ecnt++;
 }

 void tarjan(int u) {
     int v = u;
     dfn[u] = low[u] = ++dindex;
     stk[++top] = u;
     instk[u] = ;
     for(int i = head[u]; ~i; i=edge[i].next) {
         v = edge[i].v;
         if(!dfn[v]) {
             tarjan(v);
             low[u] = min(low[u], low[v]);
         }
         else if(instk[v]) low[u] = min(low[u], dfn[v]);
     }
     if(dfn[u] == low[u]) {
         bcnt++;
         do {
             v = stk[top--];
             instk[v] = ;
             belong[v] = bcnt;
         } while(v != u);
     }
 }

 LL dis(P a, P b) {
     LL l1 = a.x - b.x;
     LL l2 = a.y - b.y;
     return l1 * l1 + l2 * l2;
 }

 int main() {
     // freopen("in", "r", stdin);
     int T, _ = ;
     scanf("%d", &T);
     while(T--) {
         scanf("%d", &n);
         init();
         memset(in, , sizeof(in));
         for(int i = ; i <= n; i++) ret[i] = ;
         for(int i = ; i <= n; i++) {
             scanf("%I64d%I64d%I64d%d",&p[i].x,&p[i].y,&p[i].r,&p[i].c);
         }
         for(int i = ; i <= n; i++) {
             for(int j = ; j <= n; j++) {
                 if(i == j) continue;
                 LL d = dis(p[i], p[j]);
                 if(d <= (LL)p[i].r * p[i].r) adde(i, j);
             }
         }
         for(int i = ; i <= n; i++) {
             if(!dfn[i]) tarjan(i);
         }
         for(int i = ; i < ecnt; i++) {
             int u = edge[i].u, v = edge[i].v;
             if(belong[u] != belong[v]) in[belong[v]]++;
         }
         for(int i = ; i <= n; i++) {
             ret[belong[i]] = min(ret[belong[i]], p[i].c);
         }
         int tot = ;
         for(int i = ; i <= bcnt; i++) {
             if(!in[i]) tot += ret[i];
         }
         printf("Case #%d: ", _++);
         printf("%d\n", tot);
     }
     return ;
 }