树、图 | DFS & BFS

下面两题用DFS或BFS都可以比较顺的AC。关键是得到各个结点的深度。

1079 Total Sales of Supply Chain (25 分)

我的做法是先BFS得到各结点level，然后计算prices单价倍率数组，最后输出时再乘原始价格，尽量少些乘法次数= =。
⚠️16分和25分之间，之差float -> double，尽量用double啊

#include 
#include 
#include 
#include 

#define MAXN 100010
using namespace std;
int nn;
struct Node {
    int level;
    double amt = 0.0;
    vector children;
} tree[MAXN];

int BFS(int root) {
    queue mq;
    tree[root].level = 0;
    mq.push(root);
    int curr, lvl = 0;
    while (!mq.empty()) {
        curr = mq.front();
        mq.pop();
        lvl = tree[curr].level + 1;
        for (int i = 0; i < tree[curr].children.size(); ++i) {
            int child = tree[curr].children[i];
            tree[child].level = lvl;
            mq.push(child);
        }
    }
    return lvl;
}

int main() {
    double ori, rate;
    scanf("%d%lf%lf", &nn, &ori, &rate);
    vector retailer;
    for (int i = 0; i < nn; ++i) {
        int nch;
        scanf("%d", &nch);
        if (nch) {
            int ch;
            while (nch--) {
                scanf("%d", &ch);
                tree[i].children.emplace_back(ch);
            }
        } else {
            scanf("%lf", &tree[i].amt);
            retailer.emplace_back(i);
        }
    }
    int mlvl = BFS(0);
    vector prices;
    double rrr = 1.0, rr = 1.0 + rate / 100.0, total = 0.0;
    for (int k = 0; k <= mlvl; ++k) {
        prices.emplace_back(rrr);
        rrr *= rr;
    }
    for (int j = 0; j < retailer.size(); ++j) {
        total += tree[retailer[j]].amt * prices[tree[retailer[j]].level];
    }
    printf("%.1lf\n", total * ori);
    return 0;
}

1090 Highest Price in Supply Chain (25 分)

⚠️DFS求结点深度时，要将depth作为参数一层一层传进去，而不能仅仅把depth作为全局变量。max_depth和零售商数量可用全局变量。
⚠️最高价零售商数量：计数最大深度结点即可。

#include 
#include 
#include 

#define MAXN 100010

using namespace std;
int nn, max_depth = -1, retailer_cnt = 0;
vector nodes[MAXN];

void dfs(int root, int depth) {
    int size = nodes[root].size();
    if (size == 0) {
        if (depth > max_depth) {
            max_depth = depth;
            retailer_cnt = 1;
        } else if (depth == max_depth)
            retailer_cnt++;
    }
    for (int i = 0; i < size; ++i) {
        dfs(nodes[root][i], depth + 1);
    }
}

int main() {
    double ori_price, rate;
    scanf("%d%lf%lf", &nn, &ori_price, &rate);
    int root, father;
    for (int i = 0; i < nn; ++i) {
        scanf("%d", &father);
        if (father == -1) root = i;
        else nodes[father].emplace_back(i);
    }
    dfs(root, 0);
    printf("%.2lf %d\n", ori_price * pow(1.0 + rate / 100.0, max_depth), retailer_cnt);
    return 0;
}

1094 The Largest Generation (25 分)

#include 
#include 
#include 

#define MAXN 101
using namespace std;
vector tree[MAXN];
vector cnts;
int nn, mm, level[MAXN];

void BFS(int root) {
    queue mq;
    mq.push(root);
    level[root] = 1;
    cnts.emplace_back(1);
    while (!mq.empty()) {
        int curr = mq.front();
        mq.pop();
        for (int i = 0; i < tree[curr].size(); ++i) {
            if (cnts.size() > level[curr])
                cnts.emplace_back(0);
            mq.push(tree[curr][i]);
            level[tree[curr][i]] = level[curr] + 1;
            cnts[level[curr] + 1]++;
        }
    }
}

int main() {
    scanf("%d%d", &nn, &mm);
    cnts.emplace_back(0);
    int member, nch, child;
    for (int i = 0; i < mm; ++i) {
        scanf("%d%d", &member, &nch);
        for (int j = 0; j < nch; ++j) {
            scanf("%d", &child);
            tree[member].emplace_back(child);
        }
    }
    BFS(1);
    int Mcnt = 0, gen = -1;
    for (int i = 0; i < cnts.size(); ++i) {
        if (cnts[i] > Mcnt) {
            Mcnt = cnts[i];
            gen = i;
        }
    }
    printf("%d %d\n", Mcnt, gen);
    return 0;
}

---

1205 离考研还有15天 我好刚啊
DFS版 更简单嗷

#include 
#include 
#include 

using namespace std;
int nn, cnts[100] = {0}, M = 0;
vector gra[100];

void DFS(int root, int level) {
    cnts[level]++;
    M = max(level, M);
    for (auto item: gra[root]) {
        DFS(item, level + 1);
    }
}

int main() {
    int mm, anc = 1;
    scanf("%d%d", &nn, &mm);
    for (int i = 0; i < mm; i++) {
        int par, temp, kid;
        scanf("%d%d", &par, &temp);
        for (int j = 0; j < temp; j++) {
            scanf("%d", &kid);
            gra[par].push_back(kid);
        }
    }
    DFS(anc, 1);
    int midx = 0;
    for (int i = 1; i <= M; i++) {
        if (cnts[i] > cnts[midx])
            midx = i;
    }
    printf("%d %d\n", cnts[midx], midx);
    return 0;
}

1004 Counting Leaves (30 分)

#include 
#include 
#include 

using namespace std;
int nn, n_hasleaf;
vector nodes[101];
int level_leaf_cnt[101], Mlevel = -1;

void DFS(int root, int level) {
    Mlevel = max(Mlevel, level);
    int size = nodes[root].size();
    if (size == 0) level_leaf_cnt[level]++;
    else
        for (int i = 0; i < size; ++i) {
            DFS(nodes[root][i], level + 1);
        }
}

int main() {
    while (scanf("%d", &nn) != EOF) {
        if (nn == 0) break;
        scanf("%d", &n_hasleaf);
        for (int i = 0; i < n_hasleaf; ++i) {
            int father, cnt, child;
            scanf("%d%d", &father, &cnt);
            for (int j = 0; j < cnt; ++j) {
                scanf("%d", &child);
                nodes[father].emplace_back(child);
            }
        }
    }
    DFS(1, 0);
    for (int i = 0; i <= Mlevel; ++i) {
        printf("%d", level_leaf_cnt[i]);
        if (i < Mlevel) printf(" ");
    }
    return 0;
}