HDOJ 5137 How Many Maos Does the Guanxi Worth(最短削除点)
6373 ワード
How Many Maos Does the Guanxi Worth
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 1031 Accepted Submission(s): 353
Problem Description
"Guanxi"is a very important word in Chinese. It kind of means "relationship"or "contact". Guanxi can be based on friendship, but also can be built on money. So Chinese often say "I don't have one mao (0.1 RMB) guanxi with you."or "The guanxi between them is naked money guanxi."It is said that the Chinese society is a guanxi society, so you can see guanxi plays a very important role in many things. Here is an example. In many cities in China, the government prohibit the middle school entrance examinations in order to relief studying burden of primary school students. Because there is no clear and strict standard of entrance, someone may make their children enter good middle schools through guanxis. Boss Liu wants to send his kid to a middle school by guanxi this year. So he find out his guanxi net. Boss Liu's guanxi net consists of N people including Boss Liu and the schoolmaster. In this net, two persons who has a guanxi between them can help each other. Because Boss Liu is a big money(In Chinese English, A "big money"means one who has a lot of money) and has little friends, his guanxi net is a naked money guanxi net -- it means that if there is a guanxi between A and B and A helps B, A must get paid. Through his guanxi net, Boss Liu may ask A to help him, then A may ask B for help, and then B may ask C for help ...... If the request finally reaches the schoolmaster, Boss Liu's kid will be accepted by the middle school. Of course, all helpers including the schoolmaster are paid by Boss Liu. You hate Boss Liu and you want to undermine Boss Liu's plan. All you can do is to persuade ONE person in Boss Liu's guanxi net to reject any request. This person can be any one, but can't be Boss Liu or the schoolmaster. If you can't make Boss Liu fail, you want Boss Liu to spend as much money as possible. You should figure out that after you have done your best, how much at least must Boss Liu spend to get what he wants. Please note that if you do nothing, Boss Liu will definitely succeed.
Input
There are several test cases. For each test case: The first line contains two integers N and M. N means that there are N people in Boss Liu's guanxi net. They are numbered from 1 to N. Boss Liu is No. 1 and the schoolmaster is No. N. M means that there are M guanxis in Boss Liu's guanxi net. (3 <=N <= 30, 3 <= M <= 1000) Then M lines follow. Each line contains three integers A, B and C, meaning that there is a guanxi between A and B, and if A asks B or B asks A for help, the helper will be paid C RMB by Boss Liu. The input ends with N = 0 and M = 0. It's guaranteed that Boss Liu's request can reach the schoolmaster if you do not try to undermine his plan.
Output
For each test case, output the minimum money Boss Liu has to spend after you have done your best. If Boss Liu will fail to send his kid to the middle school, print "Inf"instead.
Sample Input
Sample Output
劉さんは自分の関係を利用して校長を見つけて、子供を良い学校に入れなければならない.各階の関係を探すには劉さんのお金がかかります.今、劉社長関係網を削除することができます.劉社長と校長のいずれかの人(1は劉社長、nは校長)を除いて、劉社長は校長を見つけることができません.劉社長が校長を見つけるのを止められないなら、劉社長の費用を最大にしなければならない.
劉さんは校長先生を見つけて、自分の費用を最小限に抑えるに違いない.図の中で点を削除して始点から終点まで連通できないようにし、任意の点を削除して始点から終点まで任意に連通すれば、任意の点を削除した場合に最短経路の値を最大にすることを探す.
注意:図の削除は、この点に接続されているすべてのエッジを削除することに等しい.(行の次の点を削除する前に、前の点を復元します)
dijkstraアルゴリズム、具体的なコードは以下の通りです.
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 1031 Accepted Submission(s): 353
Problem Description
"Guanxi"is a very important word in Chinese. It kind of means "relationship"or "contact". Guanxi can be based on friendship, but also can be built on money. So Chinese often say "I don't have one mao (0.1 RMB) guanxi with you."or "The guanxi between them is naked money guanxi."It is said that the Chinese society is a guanxi society, so you can see guanxi plays a very important role in many things. Here is an example. In many cities in China, the government prohibit the middle school entrance examinations in order to relief studying burden of primary school students. Because there is no clear and strict standard of entrance, someone may make their children enter good middle schools through guanxis. Boss Liu wants to send his kid to a middle school by guanxi this year. So he find out his guanxi net. Boss Liu's guanxi net consists of N people including Boss Liu and the schoolmaster. In this net, two persons who has a guanxi between them can help each other. Because Boss Liu is a big money(In Chinese English, A "big money"means one who has a lot of money) and has little friends, his guanxi net is a naked money guanxi net -- it means that if there is a guanxi between A and B and A helps B, A must get paid. Through his guanxi net, Boss Liu may ask A to help him, then A may ask B for help, and then B may ask C for help ...... If the request finally reaches the schoolmaster, Boss Liu's kid will be accepted by the middle school. Of course, all helpers including the schoolmaster are paid by Boss Liu. You hate Boss Liu and you want to undermine Boss Liu's plan. All you can do is to persuade ONE person in Boss Liu's guanxi net to reject any request. This person can be any one, but can't be Boss Liu or the schoolmaster. If you can't make Boss Liu fail, you want Boss Liu to spend as much money as possible. You should figure out that after you have done your best, how much at least must Boss Liu spend to get what he wants. Please note that if you do nothing, Boss Liu will definitely succeed.
Input
There are several test cases. For each test case: The first line contains two integers N and M. N means that there are N people in Boss Liu's guanxi net. They are numbered from 1 to N. Boss Liu is No. 1 and the schoolmaster is No. N. M means that there are M guanxis in Boss Liu's guanxi net. (3 <=N <= 30, 3 <= M <= 1000) Then M lines follow. Each line contains three integers A, B and C, meaning that there is a guanxi between A and B, and if A asks B or B asks A for help, the helper will be paid C RMB by Boss Liu. The input ends with N = 0 and M = 0. It's guaranteed that Boss Liu's request can reach the schoolmaster if you do not try to undermine his plan.
Output
For each test case, output the minimum money Boss Liu has to spend after you have done your best. If Boss Liu will fail to send his kid to the middle school, print "Inf"instead.
Sample Input
4 5 1 2 3 1 3 7 1 4 50 2 3 4 3 4 2 3 2 1 2 30 2 3 10 0 0
Sample Output
50 Inf
劉さんは自分の関係を利用して校長を見つけて、子供を良い学校に入れなければならない.各階の関係を探すには劉さんのお金がかかります.今、劉社長関係網を削除することができます.劉社長と校長のいずれかの人(1は劉社長、nは校長)を除いて、劉社長は校長を見つけることができません.劉社長が校長を見つけるのを止められないなら、劉社長の費用を最大にしなければならない.
劉さんは校長先生を見つけて、自分の費用を最小限に抑えるに違いない.図の中で点を削除して始点から終点まで連通できないようにし、任意の点を削除して始点から終点まで任意に連通すれば、任意の点を削除した場合に最短経路の値を最大にすることを探す.
注意:図の削除は、この点に接続されているすべてのエッジを削除することに等しい.(行の次の点を削除する前に、前の点を復元します)
dijkstraアルゴリズム、具体的なコードは以下の通りです.
#include<cstdio>
#include<cstring>
#define INF 0x3f3f3f
#define maxn 35
#define maxm 1010
int map[maxn][maxn],dis[maxn],n;
int s[maxm],e[maxm],d[maxm];
void dijkstar()
{
int i,j,min,next=1;
int visit[maxn];
for(i=1;i<=n;++i)
{
dis[i]=map[1][i];
visit[i]=0;
}
visit[1]=1;
for(i=2;i<=n;++i)
{
min=INF;
for(j=1;j<=n;++j)
{
if(!visit[j]&&min>dis[j])
{
next=j;
min=dis[j];
}
}
visit[next]=1;
for(j=1;j<=n;++j)
{
if(!visit[j]&&dis[j]>dis[next]+map[next][j])
dis[j]=dis[next]+map[next][j];
}
}
}
int main()
{
int m,i,j,a,b,c;
while(scanf("%d%d",&n,&m)&&n||m)
{
for(i=1;i<=n;++i)
{
d[i]=INF;//
for(j=1;j<=n;++j)
{
if(i==j)
map[i][j]=0;
else
map[i][j]=INF;
}
}
for(i=0;i<m;++i)
{
scanf("%d%d%d",&a,&b,&c);
if(map[a][b]>c)
map[a][b]=map[b][a]=c;
}
int sign=0;
int ans=-1;
for(i=2;i<n;++i)// 2 n-1，
{
for(j=1;j<=n;++j)//
{
if(map[i][j]!=INF&&i!=j)
{
d[j]=map[i][j];//
map[i][j]=map[j][i]=INF;//
}
}
dijkstar();
if(dis[n]==INF)//
{
sign=1;
break;
}
if(dis[n]>ans)//
ans=dis[n];
for(j=1;j<=n;++j)//
{
if(d[j]!=INF)
{
map[i][j]=map[j][i]=d[j];//
d[j]=INF;//
}
}
}
if(sign)
printf("Inf
");
else
printf("%d
",ans);
}
return 0;
}