HDU 4705 Y

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Y

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 93 Accepted Submission(s): 32

Problem Description

Sample Input

Sample Output

  
   1

   
    
     Hint
    
1. The only set is {2,3,4}.

2. Please use #pragma comment(linker, "/STACK:16777216")

Source

2013 Multi-University Training Contest 10

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zhuyuanchen520

题意：有一颗树，选出3个点。不在同一条路径上的集合数。

思路： DFS

间接法。

总的方案 - 在同一路径的。

从叶子节点开始。

假设以i为节点子树的数量分别为 sum1, sum2, sumk；（i必选）

那么对每子树选一个点。其他子树选一个点。有 sumi * (n - 1 - sumi)

然后对i的子树选一个点，i树以外选一个点。

结果会重复算了一次。所以要除上2.

#pragma comment(linker, "/STACK:16777216")
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
const int V = 100000 + 50;
const int mod = 1000000000 + 7;
vector<int> vec[V];
int n;
bool vis[V];
__int64 ans;
int dfs(int index) {
    int sum = 0;
    vis[index] = true;
    for(int i = 0; i < vec[index].size(); ++i)
        if(!vis[vec[index][i]]) {
            int temp = dfs(vec[index][i]);
            sum += temp;
            ans += (__int64) temp * (n - 1 - temp);
            //printf("%d %d %I64d\n", temp, sum, ans);
        }
    if(sum)
        ans += (__int64) sum * (n - 1 - sum);
    sum++;
    return sum;
}
int main() {
    int i, j;
    while(~scanf("%d", &n)) {
        ans = 0;
        memset(vis, false, sizeof(vis));
        for(i = 1; i < n; ++i) {
            int a, b;
            scanf("%d%d", &a, &b);
            vec[a].push_back(b);
            vec[b].push_back(a);
        }
        dfs(1);
        __int64 total = (__int64) n * (n - 1) * (n - 2) / 6;
        printf("%I64d\n", total - ans / 2);
        for(i = 1; i <= n; ++i)
            vec[i].clear();
    }
}

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#pragma comment(linker, "/STACK:16777216")
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
const int V = 100000 + 50;
const int mod = 1000000000 + 7;
int pnt[V * 2], nxt[V * 2], e, head[V * 2];
int n;
bool vis[V];
__int64 ans;
inline void addedge(int u, int v) {
    pnt[e] = v;
    nxt[e] = head[u];
    head[u] = e++;
}
int dfs(int u) {
    int sum = 0;
    vis[u] = true;
    for(int i = head[u]; i != -1; i = nxt[i])
        if(!vis[pnt[i]]) {
            int temp = dfs(pnt[i]);
            sum += temp;
            ans += (__int64) temp * (n - 1 - temp);
            //printf("%d %d %I64d\n", temp, sum, ans);
        }
    if(sum)
        ans += (__int64) sum * (n - 1 - sum);
    sum++;
    return sum;
}
int main() {
    int i, j;
    while(~scanf("%d", &n)) {
        ans = e = 0;
        for(i = 0; i <= n; ++i) {
            vis[i] = false;
            head[i] = -1;
        }
        for(i = 1; i < n; ++i) {
            int a, b;
            scanf("%d%d", &a, &b);
            addedge(a, b);
            addedge(b, a);
        }
        dfs(1);
        __int64 total = (__int64) n * (n - 1) * (n - 2) / 6;
        printf("%I64d\n", total - ans / 2);
    }
}