HDU 3072 Intelligence System （tarjan-scc缩点 + 最小树形图）

题意:

给了一个含有N<=50000个节点的有向图，图中的两点间通信付出代为经过的边权之和，但是如果这两个点之间相互可达，代价为0

问从给定的节点向其他所有的点通信，所花费的最小代价是多少？

分析:

相互可达的就是一个强联通分量,先缩点成为DAG,最小代价就是最小树形图,因为scc内部都是0,DAG无圈根据最小树形图算法思想,直接把新图最小入边权全部加起来就是ans了

代码:

//
//  Created by TaoSama on 2015-11-03
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 5e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 1e5 + 10;

int n, m;

struct Edge {
int v, nxt, c;
} edges[M];
int head
, cnt;

void add_edge(int u, int v, int c) {
edges[cnt] = (Edge) {v, head[u], c};
head[u] = cnt++;
}

int in
, id
, dfn
, low
, stk
, dfsNum, top, scc;

void tarjan(int u) {
dfn[u] = low[u] = ++dfsNum;
in[u] = true;
stk[++top] = u;
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if(!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if(in[v]) low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u]) {
++scc;
while(true) {
int v = stk[top--];
id[v] = scc;
in[v] = false;
if(v == u) break;
}
}
}

void init() {
scc = cnt = top = dfsNum = 0;
memset(head, -1, sizeof head);
memset(dfn, 0, sizeof dfn);
memset(low, 0, sizeof low);
memset(in, 0, sizeof in);
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d", &n, &m) == 2) {
init();
for(int i = 1; i <= m; ++i) {
int u, v, c; scanf("%d%d%d", &u, &v, &c);
add_edge(u, v, c);
}
for(int i = 0; i < n; ++i)
if(!dfn[i]) tarjan(i);

memset(in, 0x3f, sizeof in);
for(int i = 0; i < n; ++i) {
int u = id[i];
for(int j = head[i]; ~j; j = edges[j].nxt) {
int v = id[edges[j].v];
if(u != v) in[v] = min(in[v], edges[j].c);
}
}
int ans = 0;
//no "in" edge of root, in[root] = INF;
for(int i = 1; i <= scc; ++i)
if(in[i] != INF) ans += in[i];
printf("%d\n", ans);
}
return 0;
}