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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 1010
#define maxm 1000100
#define INF 0x3f3f3f3f
using namespace std;
int n, m;
int low[maxn];
int dfn[maxn];
int head[maxn], cnt;
int dfs_clock;
int mark;//判断图是否连通
struct node{
int u, v, w, cnt, next;
};
node edge[maxm];
void init(){
cnt = 0;
memset(head, -1, sizeof(head));
}
void add(int u, int v, int w){
edge[cnt] = {u, v, w, 0, head[u]};
head[u] = cnt++;
edge[cnt] = {v, u, w, 0, head[v]};
head[v] = cnt++;
}
void input(){
while(m--){
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
}
}
void tarjin(int u, int fa){
low[u] = dfn[u] = ++dfs_clock;
int flag = 1;
for(int i = head[u]; i != -1; i = edge[i].next){
int v = edge[i].v;
if(flag && v == fa){//去重边，重边不可能是桥
flag = 0;
continue;
}
if(!dfn[v]){
tarjin(v, u);
low[u] = min(low[u], low[v]);
if(dfn[u] < low[v])//是桥
edge[i].cnt = edge[i ^ 1].cnt = 1;
}
else
low[u] = min(low[u], dfn[v]);
}
}
void find(){
memset(dfn, 0, sizeof(dfn));
memset(low, 0, sizeof(low));
dfs_clock = 0;
tarjin(1, 1);
mark = 1;
for(int i = 1; i <= n; ++i){
if(!dfn[i]){
mark = 0;
return ;
}
}
}
void solve(){
if(!mark)
printf("0\n");
else{
int ans = INF;
for(int i = 0; i < cnt; ++i){
if(edge[i].cnt)
ans = min(ans, edge[i].w);
}
if(ans == INF)
ans = -1;
if(ans == 0)
ans = 1;
printf("%d\n", ans);
}
}
int main (){
while(scanf("%d%d", &n, &m), n || m){
init();
input();
find();
solve();
}
return 0;
}
另一种去重边的方式，对这一题来说处理时间比较慢
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 1010
#define maxm 1000100
#define INF 0x3f3f3f3f
using namespace std;
int n, m;
int low[maxn];
int dfn[maxn];
int head[maxn], cnt;
int dfs_clock;
int mark;//判断图是否连通
struct node{
int u, v, w, cnt, next, again;//again标记是不是有重边
};
node edge[maxm];
void init(){
cnt = 0;
memset(head, -1, sizeof(head));
}
void add(int u, int v, int w){
int i;
//标记重边
for(i = head[u]; i != -1; i = edge[i].next){
if(edge[i].v == v){
break;
}
}
if(i != -1)
edge[i].again = 1;
else{
edge[cnt] = {u, v, w, 0, head[u], 0};
head[u] = cnt++;
}
}
void input(){
while(m--){
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
}
void tarjin(int u, int fa){
low[u] = dfn[u] = ++dfs_clock;
for(int i = head[u]; i != -1; i = edge[i].next){
int v = edge[i].v;
if(v == fa) continue;
if(!dfn[v]){
tarjin(v, u);
low[u] = min(low[u], low[v]);
if(dfn[u] < low[v] && edge[i].again == 0)//是桥
edge[i].cnt = edge[i ^ 1].cnt = 1;
}
else
low[u] = min(low[u], dfn[v]);
}
}
void find(){
memset(dfn, 0, sizeof(dfn));
memset(low, 0, sizeof(low));
dfs_clock = 0;
tarjin(1, -1);
mark = 1;
for(int i = 1; i <= n; ++i){
if(!dfn[i]){
mark = 0;
return ;
}
}
}
void solve(){
if(!mark)
printf("0\n");
else{
int ans = INF;
for(int i = 0; i < cnt; ++i){
if(edge[i].cnt)
ans = min(ans, edge[i].w);
}
if(ans == INF)
ans = -1;
if(ans == 0)
ans = 1;
printf("%d\n", ans);
}
}
int main (){
while(scanf("%d%d", &n, &m), n || m){
init();
input();
find();
solve();
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。
HDU 4738 --Caocao's Bridges 【无向图边双联通 && 求权值最小的桥 && 模板】
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原文地址：http://blog.csdn.net/hpuhjh/article/details/47683907
