POJ 3177 Redundant Paths 边双联通分量 缩点
问最少加多少条边使得任意两点间均有2条不同的路连接。
发现，如果两个点在一个边双联通分量内，路就必定至少2条，因此缩点成树，问题转化为将树补成边双联通分量，至少要加多少条边。与POJ 3352实际上一样。
但是此题有重边，而判断割边的时候，就不能只判断父亲了，而是要判断反向边，否则重边判断就跪了。
#include <cstdio>
#include <cstring>
#include <algorithm>
#define FOR(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int N = 5005, M = 20005;
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] = 1;
dfn[u] = low[u] = ++ts;
for (int i = h[u]; i != -1; i = p[i]) {
if (i == (fa ^ 1)) 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] = 0;
belong[j] = bcc;
} while (j != u);
}
}
int main() {
int i, j, x, y, n, m, leaf;
while (scanf("%d%d", &n, &m) == 2) {
memset(instack, 0, sizeof instack);
memset(dfn, 0, sizeof dfn);
memset(h, -1, sizeof h);
memset(out, 0, sizeof out);
bcc = cnt = top = ts = leaf = 0;
while (m--) {
scanf("%d%d", &x, &y);
add(x, y); add(y, x);
}
tarjan(1, -1);
FOR(i,1,n) for (j = h[i]; j != -1; j = p[j])
if (belong[i] != belong[v[j]])
++out[belong[i]];
FOR(i,1,bcc) if (out[i] == 1) ++leaf;
printf("%d\n", (leaf + 1) / 2);
}
return 0;
}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.