hdu 4738 Caocao's Bridges(2013杭州网络赛丶神坑)

hdu 4738 Caocao's Bridges(2013杭州网络赛丶神坑)

分类： 图论2013-09-15
19:32 84人阅读 评论(2) 收藏 举报

就是求最小权值的桥。。不过有好几个坑。。。

1：原图不连通，ans=0.

2： m<=n^2 显然有重边，重边必然不是桥，处理重边直接add(u, v, INF).

3: 最小桥边权为0的时候，ans=1，至少要派一个人运炸弹。。。

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#include<iostream>

#include<algorithm>

#include<vector>

#include<string>

#include<queue>

#include<stack>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<fstream>

#include<sstream>

#include<map>

#include<set>

using namespace std;

const int N = 1111;

const int INF = 100000000;

int n, m;

int pre
, low
, dfs_clock;

int bccno
, vis
, wi

, g

;

struct Edge

{

//flag = 1 ->bridge

int from, to, w, flag;

};

vector<int> G
;

vector<Edge> edges;

//add bidir edge

void addedge(int u, int v, int w)

{

edges.push_back((Edge){u, v, w, 0});

edges.push_back((Edge){v, u, w, 0});

int nima = edges.size();

G[u].push_back(nima-2);

G[v].push_back(nima-1);

}

int dfs(int u, int fa)

{

int lowu = pre[u] = ++dfs_clock;

int sz = G[u].size();

for(int i=0; i<sz; i++)

{

int v = edges[G[u][i]].to;

if(!pre[v])

{

int lowv = dfs(v, u);

lowu = min(lowu, lowv);

if(lowv > pre[u]) edges[G[u][i]].flag = 1, edges[G[u][i]^1].flag = 1;

}

else if(pre[v] < pre[u] && v != fa) lowu = min(lowu, pre[v]);

}

return low[u] = lowu;

}

void dfs1(int u)

{

vis[u] = 1;

int sz = G[u].size();

for(int i=0; i<sz; i++)

{

Edge e = edges[G[u][i]];

int v = e.to;

if(!vis[v]) dfs1(v);

}

}

int u, v, w;

int main()

{

while(scanf("%d%d", &n, &m), n+m)

{

for(int i=0; i<n+1; i++) G[i].clear();

edges.clear();

memset(g, 0, sizeof(g));

for(int i=0; i<m; i++)

{

scanf("%d%d%d", &u, &v, &w); u--; v--;

g[u][v]++; //BIANSHU

g[v][u]++;

wi[u][v] = w;

}

//CHONG BIAN

for(int i=0; i<n; i++)

for(int j=0; j<n; j++)

{

if(g[i][j] == 1) addedge(i, j, wi[i][j]);

else if(g[i][j] > 1) addedge(i, j, INF);

}

//BU LIANTONG

memset(vis, 0, sizeof(vis));

dfs1(0);

bool flag = 0;

for(int i=0; i<n; i++)

{

if(!vis[i])

{

flag = 1;

break;

}

}

if(flag)

{

puts("0");

continue;

}

//QIAO

memset(pre, 0, sizeof(pre));

dfs_clock = 0;

for(int i=0; i<n; i++) if(!pre[i]) dfs(i, -1);

int ans = INF;

int sz = edges.size();

for(int i=0; i<sz; i++) if(edges[i].flag == 1) ans = min(ans, edges[i].w);

if(ans == INF) puts("-1");

else printf("%d\n", ans == 0 ? 1 : ans); //0 -> 1

}

return 0;

}