HDU 1827 Summer Holiday
题意:中文题
思路:强连通缩点,每一个点的权值为强连通中最小值,然后入度为0的点就是答案
代码:
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <stack>
using namespace std; const int N = 1005;
const int INF = 0x3f3f3f3f; int n, m, val[N];
vector<int> g[N];
int pre[N], lowlink[N], dfs_clock, sccno[N], scc_cnt, scc_val[N];
stack<int> S; void dfs(int u) {
pre[u] = lowlink[u] = ++dfs_clock;
S.push(u);
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (!pre[v]) {
dfs(v);
lowlink[u] = min(lowlink[u], lowlink[v]);
} else if (!sccno[v]) lowlink[u] = min(lowlink[u], pre[v]);
}
if (pre[u] == lowlink[u]) {
scc_cnt++;
scc_val[scc_cnt] = INF;
while (1) {
int x = S.top(); S.pop();
scc_val[scc_cnt] = min(scc_val[scc_cnt], val[x]);
sccno[x] = scc_cnt;
if (x == u) break;
}
}
} void find_scc() {
dfs_clock = scc_cnt = 0;
memset(pre, 0, sizeof(pre));
memset(sccno, 0, sizeof(sccno));
for (int i = 1; i <= n; i++)
if (!pre[i]) dfs(i);
} int in[N]; int main() {
while (~scanf("%d%d", &n, &m)) {
for (int i = 1; i <= n; i++) {
g[i].clear();
scanf("%d", &val[i]);
}
int u, v;
while (m--) {
scanf("%d%d", &u, &v);
g[u].push_back(v);
}
find_scc();
memset(in, 0, sizeof(in));
for (int u = 1; u <= n; u++) {
for (int j = 0; j < g[u].size(); j++) {
int v = g[u][j];
if (sccno[u] != sccno[v])
in[sccno[v]]++;
}
}
int ans = 0, cnt = 0;
for (int i = 1; i <= scc_cnt; i++)
if (!in[i]) {
ans += scc_val[i];
cnt++;
}
printf("%d %d\n", cnt, ans);
}
return 0;
}