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分支限界法算法设计.cpp

@@ -0,0 +1,55 @@
+#include <iostream>
+#include <vector>
+#include <queue>
+#include <cmath>
+using namespace std;
+
+const int N = (1 << 20) + 10;
+int n,m;
+int v[N],w[N];
+int res = -1;
+int s[N]; // s[i]存储的是从第1件物品到第i件物品的价值总和
+
+struct good{
+    int idx,c,r,tv; // idx表示选法下标，c表示该选法的当前总价值，r表示当前选法的剩余总价值，tv表示该选法的当前总体积
+    bool operator > (const good& W) const{
+        return W.c + W.r > c + r;
+    }
+}goods[N];
+
+int bfs(){
+    goods[1] = {1,0,0,0};
+priority_queue<good,vector<good>,greater<good>> q;
+    q.push(goods[1]);
+
+    while(q.size()){
+        auto t = q.top();
+//        cout << t.idx << endl;
+        q.pop();
+        int idx = t.idx << 1;
+        goods[idx] = {idx,goods[t.idx].c,s[n] - s[(int)log2(idx)],goods[t.idx].tv};
+        goods[idx + 1] = {idx + 1,goods[t.idx].c + w[(int)log2(idx)],s[n] - s[(int)log2(idx)],goods[t.idx].tv + v[(int)log2(idx)]};
+        if((int)log2(t.idx) == n) { // 假如已经是子节点，则更新答案
+            res = max(res,t.c);
+            continue;
+        }
+
+        if(goods[idx].tv <= m && goods[idx].c + goods[idx].r > res) q.push(goods[idx]); // 假如当前选法的总体积不超过背包容量，且当前价值+剩余价值 > 当前最优解，则装入背包
+        if(goods[idx + 1].tv <= m && goods[idx + 1].c + goods[idx + 1].r > res) q.push(goods[idx + 1]);
+    }
+
+    return res;
+}
+
+int main(){
+    cin >> n >> m;
+    for(int i = 1; i <= n; i ++) cin >> v[i] >> w[i],s[i] = s[i - 1] + w[i];
+//
+//    for(int i = 2; i < 1 << n + 1; i ++){
+//        goods[i] = {i,goods[i >> 1].c + (i&1)*w[(int)log2(i)],s[n] - s[(int)log2(i)],goods[i >> 1].tv + (i&1)*v[(int)log2(i)]};
+//    }
+//
+//    for(int i = 1; i < 1 << n + 1; i ++) printf("i = %d,c[i] = %d,r[i] = %d,tv[i] = %d\n",goods[i].idx,goods[i].c,goods[i].r,goods[i].tv);
+    cout << bfs() << endl;
+    return 0;
+}