hdu 4738 Caocao's Bridges 图--桥的判断模板
Caocao's Bridges
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1231 Accepted Submission(s): 478
Problem Description
Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn't give up. Caocao's army still was not good at water battles, so he came up with another idea. He built many islands in the Changjiang river, and based on those islands, Caocao's army could easily attack Zhou Yu's troop. Caocao also built bridges connecting islands. If all islands were connected by bridges, Caocao's army could be deployed very conveniently among those islands. Zhou Yu couldn't stand with that, so he wanted to destroy some Caocao's bridges so one or more islands would be seperated from other islands. But Zhou Yu had only one bomb which was left by Zhuge Liang, so he could only destroy one bridge. Zhou Yu must send someone carrying the bomb to destroy the bridge. There might be guards on bridges. The soldier number of the bombing team couldn't be less than the guard number of a bridge, or the mission would fail. Please figure out as least how many soldiers Zhou Yu have to sent to complete the island seperating mission.
Input
There are no more than 12 test cases.
In each test case:
The first line contains two integers, N and M, meaning that there are N islands and M bridges. All the islands are numbered from 1 to N. ( 2 <= N <= 1000, 0 < M <= N 2 )
Next M lines describes M bridges. Each line contains three integers U,V and W, meaning that there is a bridge connecting island U and island V, and there are W guards on that bridge. ( U ≠ V and 0 <= W <= 10,000 )
The input ends with N = 0 and M = 0.
In each test case:
The first line contains two integers, N and M, meaning that there are N islands and M bridges. All the islands are numbered from 1 to N. ( 2 <= N <= 1000, 0 < M <= N 2 )
Next M lines describes M bridges. Each line contains three integers U,V and W, meaning that there is a bridge connecting island U and island V, and there are W guards on that bridge. ( U ≠ V and 0 <= W <= 10,000 )
The input ends with N = 0 and M = 0.
Output
For each test case, print the minimum soldier number Zhou Yu had to send to complete the mission. If Zhou Yu couldn't succeed any way, print -1 instead.
Sample Input
3 3
1 2 7
2 3 4
3 1 4
3 2
1 2 7
2 3 4
0 0
Sample Output
-1
4
Source
图的连通性问题~之求桥模板
#include "stdio.h" //本人觉得两点之间可能有多条边，这样的话，邻接矩阵就没法存了，转邻接表~
#include "string.h"
#define N 1005
#define INF 0x3fffffff
struct node
{
int x,y;
int weight;
int next;
}edge[4*N*N];
int idx,head[N];
int set[N]; //并查集用
int root;
bool mark[N],visit[N];
int low[N],dfn[N];
int stackk[4*N*N],num;
int MIN(int a,int b){ return a<b?a:b; }
void Init(){ idx=0; memset(head,-1,sizeof(head)); }
int find(int x){ return set[x]==x?set[x]:set[x]=find(set[x]); }
void Add(int x,int y,int weight)
{
edge[idx].x = x;
edge[idx].y = y;
edge[idx].weight = weight;
edge[idx].next = head[x];
head[x] = idx++;
}
void Union(int x,int y)
{
int fa = find(x);
int fb = find(y);
if(fa!=fb)
set[fa] = fb;
}
void DFS(int x,int times,int edge_father)
{
int i,y;
int child=0;
visit[x] = true;
low[x] = dfn[x] = times;
for(i=head[x]; i!=-1; i=edge[i].next)
{
y=edge[i].y;
if(!visit[y])
{
child++;
DFS(y,times+1,i);
low[x] = MIN(low[x],low[y]);
//if(x==root && child==2) mark[root] = true; //记录顶点是否为割顶
//if(x!=root && low[y]>=dfn[x]) mark[x] = true; //记录顶点是否为割顶
if(low[y]>dfn[x]) stackk[num++] = i; //若边i为桥，存入stackk[];
}
else if(edge_father!=-1 && i!=(edge_father^1)) //若当前边不为原来他father到他的边，更新low[x];
low[x] = MIN(low[x],dfn[y]);
}
}
int solve(int n)
{
int i,ans = INF;
num = 0;
memset(visit,false,sizeof(visit)); //标记点是否已访问
//memset(mark,false,sizeof(mark));
root = 1;
int times = 1;
DFS(root,times,-1);
for(i=0; i<num; ++i)
ans = MIN(ans,edge[stackk[i]].weight);
if(ans==INF) ans=-1; //不存在桥，ans=-1；
if(ans==0) ans++; //如果桥权值是零，则最少要派一人去炸桥
return ans;
}
int main()
{
int i;
int n,m;
int x,y,k;
while(scanf("%d %d",&n,&m),n&&m)
{
Init();
for(i=1; i<=n; ++i) set[i] = i; //并查集用
while(m--)
{
scanf("%d %d %d",&x,&y,&k);
if(x==y) continue;
Add(x,y,k);
Add(y,x,k);
Union(x,y); //合并两点
}
bool flag = true;
for(i=1; i<=n; ++i) //有一个点不连通，则flag为false;
{
if(find(i) != find(1))
flag = false;
}
if(!flag){ printf("0\n"); continue; }
printf("%d\n",solve(n));
}
return 0;
}
//后加上的代码~~
#include "stdio.h"
#include "string.h"
#define N 1010
#define INF 0x3fffffff
struct node
{
int x,y;
bool visit;
int weight;
int next;
}edge[2*N*N];
int idx,head[N];
int n,m;
int time;
int low[N],dfn[N];
bool mark[N];
int st[2*N*N],num; //存割边的编号
int MIN(int x,int y){ return x<y?x:y; }
void Init(){idx=0; memset(head,-1,sizeof(head)); }
void Add(int x,int y,int k)
{
edge[idx].x = x;
edge[idx].y = y;
edge[idx].visit = false; //该条边未被访问
edge[idx].weight = k;
edge[idx].next = head[x];
head[x] = idx++;
}
int set[N];
int find(int x);
void Union(int x,int y);
void DFS(int x)
{
int i,y;
low[x] = dfn[x] = ++time;
for(i=head[x]; i!=-1; i=edge[i].next)
{
y = edge[i].y;
if(edge[i].visit) continue; //该边已经访问过，continue；
edge[i].visit = edge[i^1].visit = true;
if(!dfn[y])
{
DFS(y);
low[x] = MIN(low[x],low[y]);
if(low[y]>dfn[x])
st[num++] = i;
}
else
low[x] = MIN(low[x],dfn[y]);
}
}
int Solve()
{
int ans = INF;
int i,j;
time = 0;
num = 0;
memset(dfn,0,sizeof(dfn));
DFS(1);
for(i=0; i<num; ++i)
{
if(ans>edge[st[i]].weight)
ans = edge[st[i]].weight;
}
if(ans==INF) ans=-1;
if(ans==0) ans=1;
return ans;
}
int main()
{
bool flag;
int i,j;
int x,y,k;
while(scanf("%d %d",&n,&m),n||m)
{
Init();
for(i=1; i<=n; ++i) set[i] = i;
for(i=1; i<=m; ++i)
{
scanf("%d %d %d",&x,&y,&k);
Add(x,y,k);
Add(y,x,k);
Union(x,y);
}
flag = true;
for(i=1; i<=n; ++i)
{
if(find(i)!=find(1))
flag = false;
}
if(!flag) { printf("0\n"); continue; } //图不连通，派0个人去
printf("%d\n",Solve());
}
return 0;
}
int find(int x)
{
return set[x]==x?set[x]:set[x]=find(set[x]);
}
void Union(int x,int y)
{
int fa = find(x);
int fb = find(y);
if(fa!=fb)
set[fa] = fb;
}