POJ 3177 Redundant Paths 邊雙聯通分量 縮點

問最少加多少條邊使得任意兩點間均有2條不同的路連接配接。

發現，如果兩個點在一個邊雙聯通分量内，路就必定至少2條，是以縮點成樹，問題轉化為将樹補成邊雙聯通分量，至少要加多少條邊。與POJ 3352實際上一樣。

但是此題有重邊，而判斷割邊的時候，就不能隻判斷父親了，而是要判斷反向邊，否則重邊判斷就跪了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define FOR(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int N = , M = ;
int dfn[N], low[N], belong[N], out[N], bcc;
int stack[N], instack[N], top, ts;
int h[N], p[M], v[M], cnt;

void add(int x, int y) {
    p[cnt] = h[x]; v[cnt] = y; h[x] = cnt++;
}

void tarjan(int u, int fa) {
    stack[++top] = u;
    instack[u] = ;
    dfn[u] = low[u] = ++ts;
    for (int i = h[u]; i != -; i = p[i]) {
        if (i == (fa ^ )) continue;
        if (!dfn[v[i]]) {
            tarjan(v[i], i);
            low[u] = min(low[u], low[v[i]]);
        } else if (instack[v[i]])
            low[u] = min(low[u], dfn[v[i]]);
    }
    if (dfn[u] == low[u]) {
        ++bcc; int j;
        do {
            j = stack[top--];
            instack[j] = ;
            belong[j] = bcc;
        } while (j != u);
    }
}

int main() {
    int i, j, x, y, n, m, leaf;
    while (scanf("%d%d", &n, &m) == ) {
        memset(instack, , sizeof instack);
        memset(dfn, , sizeof dfn);
        memset(h, -, sizeof h);
        memset(out, , sizeof out);
        bcc = cnt = top = ts = leaf = ;
        while (m--) {
            scanf("%d%d", &x, &y);
            add(x, y); add(y, x);
        }
        tarjan(, -);
        FOR(i,,n) for (j = h[i]; j != -; j = p[j])
            if (belong[i] != belong[v[j]])
                ++out[belong[i]];
        FOR(i,,bcc) if (out[i] == ) ++leaf;
        printf("%d\n", (leaf + ) / );
    }
    return ;
}
           

Redundant Paths

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 11561 Accepted: 4966

Description

In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another.

Given a description of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way.

There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.

Input

Line 1: Two space-separated integers: F and R

Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.

Output

Line 1: A single integer that is the number of new paths that must be built.

Sample Input

7 7

1 2

2 3

3 4

2 5

4 5

5 6

5 7

Sample Output

2

Hint

Explanation of the sample:

One visualization of the paths is: 1 2 3 +—+—+ | | | | 6 +—+—+ 4 / 5 / / 7 + Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions. 1 2 3 +—+—+

| |

| |

6 +—+—+ 4

/ 5 :

/ :

/ :

7 + - - - -

Check some of the routes:

1 – 2: 1 –> 2 and 1 –> 6 –> 5 –> 2

1 – 4: 1 –> 2 –> 3 –> 4 and 1 –> 6 –> 5 –> 4

3 – 7: 3 –> 4 –> 7 and 3 –> 2 –> 5 –> 7

Every pair of fields is, in fact, connected by two routes.

It’s possible that adding some other path will also solve the problem (like one from 6 to 7). Adding two paths, however, is the minimum.