Codeforces Round #250 (Div. 2)

Codeforces Round #250 (Div. 2)

题目链接

A：水题。暴力推断有没有3个两倍长或两倍短的字符串。注意假设有多个答案也是属于C的情况
B：先打出lowbit的表，然后从大到小去组合就可以
C：贪心，权值大的点优先拿掉
D：并查集+贪心。先把边都按权值从大到小排序，然后加边，该边能够被用到的次数为左边集合个数乘上右边集合个数，最后答案在除以C2n就可以

代码：
A：

#include <stdio.h>
#include <string.h>

char str[4][105];

int main() {
    int i, j, flag = 0;
    char ans = 'C';
    for (i = 0; i < 4; i++)
        scanf("%s", str[i]);
    for (i = 0; i < 4; i++) {
        int s = 0, l = 0;
        for (j = 0; j < 4; j++) {
            if (i == j) continue;
            int a = strlen(str[i]) - 2;
            int b = strlen(str[j]) - 2;
            if (a * 2 <= b)
                s++;
            if (a >= b * 2)
                l++;                
        }
        if (s == 3 || l == 3) {
            ans = i + 'A';
            flag++;
        }
    }
    if (flag != 1) ans = 'C';
    printf("%c
", ans);
    return 0;
}

B:

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int N = 100001;
int sum, limit, ans[N], an = 0;
int i, j;
struct Low {
    int lowbit;
    int num;
} l[N];

bool cmp(Low a, Low b) {
    return a.lowbit > b.lowbit;
}

int main() {
    scanf("%d%d", &sum, &limit);
    for (i = 1; i < N; i++) {
        l[i].num = i;
        l[i].lowbit = (i&(-i));
    }
    sort(l + 1, l + N, cmp);
    for (i = 1; i < N; i++) {
        if (l[i].num <= limit) {
            if (sum >= l[i].lowbit) {
                ans[an++] = l[i].num;
                sum -= l[i].lowbit;
            }
        }
        if (sum == 0) break;
    }
    if (sum != 0) printf("-1
"); 
    else {
        printf("%d
", an);
        for (int i = 0; i < an - 1; i++)
            printf("%d ", ans[i]);
        printf("%d
", ans[an - 1]);
    }
    return 0;
}

C:

#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
#define max(a,b) ((a)>(b)?
(a):(b))
using namespace std;

const int N = 1005;
int n, m, vis[N], val[N];
vector<int> g[N];
struct Node {
    int value;
    int id;
} node[N];

bool cmp(Node a, Node b) {
    return a.value > b.value;
}

int main() {
    int i, j;
    scanf("%d%d", &n, &m);
    for (i = 1; i <= n; i++) {
        scanf("%d", &node[i].value);
        node[i].id = i;
        val[i] = node[i].value;
    }
    int u, v;
    while (m--) {
        scanf("%d%d", &u, &v);
        g[u].push_back(v);
        g[v].push_back(u);
    }
    int ans = 0;
    sort(node + 1, node + 1 + n, cmp);
    for (i = 1; i <= n; i++) {
        int u = node[i].id;
        vis[u] = 1;
        for (j = 0; j < g[u].size(); j++) {
            int v = g[u][j];
            if (vis[v]) continue;
            ans += val[v];
        }
    }
    printf("%d
", ans);
    return 0;
}

D:

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int N = 100005;
int n, m, node[N], parent[N], sum[N];

struct Edge {
    int u, v, w;
    Edge(int u = 0, int v = 0, int w = 0) {
    this->u = u; this->v = v; this->w = w;
    }
} e[N];

bool cmp(Edge a, Edge b) {
    return a.w > b.w;
}

int find(int x) {
    if (x == parent[x]) return x;
    return parent[x] = find(parent[x]);
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
    parent[i] = i;
    sum[i] = 1;
    scanf("%d", &node[i]);
    }
    int u, v;
    for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    e[i] = Edge(u, v, min(node[u], node[v]));
    }
    sort(e, e + m, cmp);
    double ans = 0;
    for (int i = 0; i < m; i++) {
    int pa = find(e[i].u);
    int pb = find(e[i].v);
    if (pa != pb) {
        ans += (double)e[i].w * sum[pa] * sum[pb];
        parent[pb] = pa;
        sum[pa] += sum[pb];
    }
    }
    printf("%.6lf
", ans * 2 / (n * 1.0 * (n - 1)));
    return 0;
}