I-caocao ' s bridges-hdu 4738 (for bridge)

Test instructions: There are some bridge links between Cao's boat, and now Zhou Yu wants to divide the connected ship into two parts, but he can only blow up a bridge, and there are soldiers guarding each bridge, asking how many soldiers he needs at least to blow up the bridge if not, output '- 1 ' NOTE: This problem has several pits, the first input bridge guard is 0 words also have to platoon a soldier if the first bridge is disconnected, you don't have to send the soldiers, direct output 0***************************************************** #include <stdio.h>
#include <string.h>
#include <algorithm>
usingnamespaceStd

ConstintMAXN = 1e3+5;
Const intoo = 1e9;

structedge{intV, used, Val, next;} e[maxn*maxn*2];
intHEAD[MAXN], CNT;
voidAddedge (intUintVintVal
{
if(val = =0) val =1;
E[CNT].V = v;
e[cnt].used =false;
E[cnt].val = val;
E[cnt].next = Head[u];
Head[u] = cnt++;
}

intLOW[MAXN], DFN[MAXN], Index;
intMin;

voidInIt (intN
{
CNT = Index =0;
Min = oo;
for(intI=0; i<=n; i++)
{
Head[i] =-1;
Dfn[i] =0;
}
}
voidTarjan (intU
{
Dfn[u] = low[u] = ++index;

for(intJ=head[u]; j!=-1; J=e[j].next)
{
if(e[j].used = =false)
{
intv = e[j].v;

e[j].used = e[j^1].used =true;

if(!dfn[v])
{
Tarjan (v);
Low[u] = min (Low[u], low[v]);

if(Low[v] > Dfn[u])
min = min (min, e[j].val);
}
Else
Low[u] = min (Low[u], dfn[v]);
}
}
}

intMain ()
{
intN, M;

while(SCANF ("%d%d", &n, &m), n+m)
{
intU, V, Val;

InIt (N);

while(m--)
{
scanf"%d%d%d", &u, &v, &val);
Addedge (U, V, Val);
Addedge (V, U, Val);
}

Tarjan (1);

for(intI=1; i<=n; i++)
{
if(Dfn[i] = =0)
Min =0;
}

if(Min = = oo)
printf"-1\n");
Else
printf"%d\n", Min);
}

return0; 

}

I-caocao ' s bridges-hdu 4738 (for bridge)