zoj 3583 simple path

Simple Path

Time Limit: 2 Seconds Memory Limit: 65536 KB

A path with no repeated vertices of an undirected graph is called a simple path. Given an undirected graph and two verteices S and D, return the number of vertics which don't lie on any
simple paths between S and D.

Input

The input contains multiple test cases.
Each case starts with a line of four integers, N(1 < N ≤ 100), M(1 ≤ M ≤ N(N - 1) / 2), S(0 ≤ S < N), D(0
≤ D < N). N is the number of vertices, M is the number of edges, S and D are two different vertices. Then M lines follow, each line contains two different integers A(0
≤ A < N) and B(0 ≤ B < N), which represents an edge of the graph. It's ensure that there is at least one simple path between S and D.

Output

Output the number of such vertics, one line per case.

Sample Input

4 3 0 2
0 1
1 2
1 3
4 4 0 2
0 1
1 2
1 3
2 3

Sample Output

1
0

Author: LIU, Yaoting

Contest: ZOJ 10th Anniversary Contest

枚举所有点，把这个点去掉，从source和dest各一次bfs，标记即不和source连通也不和dest连通的点，标记的点的总数就是answer

#include <cstdio>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>

#ifdef _DEBUG
#define debug_printf(...) printf(__VA_ARGS__)
#else
#define debug_printf(...) 0
#endif

const int MAXN = 100;

std::vector<std::vector<int> > adj;

bool hidden[MAXN];

void bfs(int src, int d[])
{
memset(d, -1, sizeof(*d) * MAXN);

d[src] = 0;

std::queue<int> que;
que.push(src);

while (! que.empty()) {

int v = que.front();
que.pop();

if (hidden[v]) {
continue;
}

assert(0 <= v && v < adj.size());

for (int i=0; i<adj[v].size(); ++i) {
int a = adj[v][i];
if (d[a] == -1 && !hidden[a]) {
d[a] = d[v] + 1;
que.push(a);
}
}
}
}

void solve()
{
int vertices;
int edges;
int src;
int dest;

if (scanf("%d%d%d%d", &vertices, &edges, &src, &dest) == EOF) {
exit(0);
}

adj.clear();
adj.resize(vertices);

for (int i=0; i<edges; ++i) {
int a = 0, b = 0;
scanf("%d%d", &a, &b);
assert(0 <= a && a < adj.size());
assert(0 <= b && b < adj.size());

if (std::find(adj[a].begin(), adj[a].end(), b) == adj[a].end()) {
adj[a].push_back(b);
}

if (std::find(adj[b].begin(), adj[b].end(), a) == adj[b].end()) {
adj[b].push_back(a);
}
}

bool onceUncovered[MAXN];
memset(onceUncovered, 0, sizeof(onceUncovered));

memset(hidden, 0, sizeof(hidden));

int d[MAXN];
int e[MAXN];

for (int i=0; i<vertices; ++i) {
if (i == dest || i == src) {
//continue;
}

hidden[i] = true;

bfs(src, d);
bfs(dest, e);

for (int v=0; v<vertices; ++v) {
if (d[v] == -1 && e[v] == -1 && v != i && v != src && v != dest) {
onceUncovered[v] = true;
}
}

hidden[i] = false;
}

int answer = 0;
for (int i=0; i<vertices; ++i) {
if (onceUncovered[i] && i != src && i != dest) {
++answer;
}
}

printf("%d\n", answer);
}

int main()
{
while(true) {
solve();
}
}