BC73 div2 Rikka with Graph 删除边，最多能生成几棵树 HDU 5631

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 308 Accepted Submission(s): 147



Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has a non-direct graph with n vertices and n+1 edges. Rikka can choose some of the edges (at least one) and delete them from the graph.

Yuta wants to know the number of the ways to choose the edges in order to make the remaining graph connected.

It is too difficult for Rikka. Can you help her?

Input

The first line contains a number T(T≤30)——The
number of the testcases.

For each testcase, the first line contains a number n(n≤100).

Then n+1 lines follow. Each line contains two numbers u,v ,
which means there is an edge between u and v.

Output

For each testcase, print a single number.

Sample Input

1
3
1 2
2 3
3 1
1 3

Sample Output

9

题目大意：

n个地点，n+1条路，每次删除其中几条边，但是这些地点仍然连接着。问，删除的方法有几种（删除一条或者一条以上）

n - 1 为生成了一棵树的依据，所以最多只能删除两条边

枚举，每次重连即可。最大为O(n^3)。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<string>
using namespace std;

const int maxn = 300 + 10;
struct edge{
int u, v;
};
int n;
edge ed[maxn];
int par[maxn], level[maxn];

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

bool unite(int x, int y){
x = find(x);
y = find(y);
if (x == y) return false;
if (level[x] > level[y]){
par[y] = x;
}
else {
par[x] = y;
if (level[x] == level[y]) level[x]++;
}
return true;
}

void init(){
for (int i = 0; i < n + 1; i++){
par[i] = i;
level[i] = 0;
}
}

void solve(){
int cnt = 0, tmp;
for (int i = 0; i < n + 1; i++){
for (int j = i; j < n + 1; j++){
init();
tmp = 0;
for (int k = 0; k < n + 1; k++){
if (k != i && k != j){
int flag = unite(ed[k].u, ed[k].v);
if (flag == 1) tmp++;
//			printf("tmp = %d\n", tmp);
}
}
//	printf("\n");
if (tmp == n - 1){
cnt ++;
}
}
}
printf("%d\n", cnt);
}

int main(){
int t;
scanf("%d", &t);
while(t--){
scanf("%d", &n);
for (int i = 0; i < n + 1; i++){
scanf("%d%d", &ed[i].u, &ed[i].v);
}
solve();
}
return 0;
}
</string></queue></cstring></algorithm></cstdio>