Caocao's Bridges
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3000 Accepted Submission(s):
953
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.
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 <= N2 )
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.
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
题意：曹操在长江上建造了n个岛屿，这些岛屿之间要用桥连接起来，每座桥上都派兵驻守，现在周瑜手中有一个炸弹，他要派人去炸掉其中一座桥，他派去的人数必须大于这条桥上驻守的人数，问他最少要派多少人去
注意：1、如果图本来就不连通 不需要派人输出0
2、如果桥上没有人把守，派去一个人即可
3、如果炸掉桥还不能断开所有岛屿之间的的连接，输出-1
#include<stdio.h>
#include<string.h>
#include<stack>
#include<algorithm>
#include<vector>
#define MAX 1010
#define MAXM 1000100
#define INF 0x7ffffff
using namespace std;
int n,m,mark;//mark记录图是否联通
int head[MAX],ans;
int low[MAX],dfn[MAX];
int dfsclock,dcccnt;
struct node
{
int beg,end,val,next;
int cnt;//记录桥是否存在
}edge[MAXM];
void init()
{
ans=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
edge[ans].beg=u;
edge[ans].end=v;
edge[ans].val=w;
edge[ans].cnt=0;//初始化为0表示没有桥
edge[ans].next=head[u];
head[u]=ans++;
}
void getmap()
{
int a,b,c;
while(m--)
{
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);
add(b,a,c);
}
}
void tarjan(int u,int fa)
{
int v;
low[u]=dfn[u]=++dfsclock;
int flag=1;
for(int i=head[u];i!=-1;i=edge[i].next)
{
v=edge[i].end;
if(flag&&v==fa)//判断重边
{
flag=0;
continue;
}
if(!dfn[v])
{
tarjan(v,u);
low[u]=min(low[u],low[v]);
if(dfn[u]<low[v])//是桥
edge[i].cnt=edge[i^1].cnt=1; //标记这条边是桥edge[i^1].cnt意思是这条边的反向边
}
else
low[u]=min(low[u],dfn[v]);
}
}
void find()
{
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
dfsclock=0;
tarjan(1,-1);
mark=1;
for(int i=1;i<=n;i++)//遍历所有点
{
if(!dfn[i])//如果点没被搜索到证明这个图不连通
{
mark=0;
return ;
}
}
}
void solve()
{
if(!mark)//图不连通 不需要派人 直接输出0
printf("0\n");
else
{
int ant=INF;
for(int i=0;i<ans;i++)
{
if(edge[i].cnt)
ant=min(ant,edge[i].val);//寻找人数最少把守的桥
}
if(ant==INF)//如果不能阻断各岛屿之间的联系
ant=-1;
if(ant==0)//如果桥上的人数为0 派一个人
ant=1;
printf("%d\n",ant);
}
}
int main()
{
while(scanf("%d%d",&n,&m),n|m)
{
init();
getmap();
find();
solve();
}
return 0;
}