'commit'

TangDou/AcWing/SSC/P3387.cpp

@@ -0,0 +1,87 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 100010, M = 2000010; // 因为要建新图，两倍的边
+int n, m;                          // 点数、边数
+int dfn[N], low[N], ts;
+int stk[N], top, in_stk[N];
+int id[N], scc_cnt, sz[N];
+int f[N];
+int h[N], hs[N], e[M], ne[M], idx; // h: 原图；hs: 新图
+void add(int h[], int a, int b) {
+    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
+}
+int p[N]; //原图的点权
+int q[N]; //新图的点权
+
+// tarjan求强连通分量
+void tarjan(int u) {
+    dfn[u] = low[u] = ++ts;
+    stk[++top] = u;
+    in_stk[u] = true;
+    for (int i = h[u]; ~i; i = ne[i]) {
+        int j = e[i];
+        if (!dfn[j]) {
+            tarjan(j);
+            low[u] = min(low[u], low[j]);
+        } else if (in_stk[j])
+            low[u] = min(low[u], dfn[j]);
+    }
+    if (dfn[u] == low[u]) {
+        ++scc_cnt;
+        int x;
+        do {
+            x = stk[top--];
+            in_stk[x] = false;
+            id[x] = scc_cnt;
+            sz[scc_cnt]++;
+            // scc_cnt:强连通块的编号
+            q[scc_cnt] += p[x]; //叠加点权，生成连通块的总点权
+        } while (x != u);
+    }
+}
+
+int main() {
+    memset(h, -1, sizeof h);   //原图
+    memset(hs, -1, sizeof hs); //新图
+
+    scanf("%d %d", &n, &m);
+    //读入点权
+    for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
+    for (int i = 1; i <= m; i++) {
+        int a, b;
+        scanf("%d %d", &a, &b);
+        add(h, a, b);
+    }
+    //利用强连通分量，缩点，生成DAG
+    for (int i = 1; i <= n; i++)
+        if (!dfn[i]) tarjan(i);
+
+    // (2) 缩点，建图
+    for (int u = 1; u <= n; u++)
+        for (int i = h[u]; ~i; i = ne[i]) {
+            int j = e[i];
+            int a = id[u], b = id[j];
+            if (a != b)        //去重边
+                add(hs, a, b); //加入到新图
+        }
+
+    // (3) 根据拓扑序遍历DAG，从scc_cnt向前遍历自然满足拓扑序
+    for (int u = scc_cnt; u; u--) {
+        // base case 递推起点
+        if (!f[u]) f[u] = q[u];
+
+        for (int i = hs[u]; ~i; i = ne[i]) {
+            int j = e[i]; // 边(i, j)
+            if (f[j] < f[u] + q[j]) f[j] = f[u] + q[j];
+        }
+    }
+
+    // (4) 求解答案
+    int res = 0;
+    for (int i = 1; i <= scc_cnt; i++) res = max(res, f[i]);
+
+    //输出
+    printf("%d\n", res);
+    return 0;
+}