hdoj 4738 Caocao's Bridges【雙連通分量求橋】

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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 DescriptionCaocao 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. 

 

InputThere 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. 

 

OutputFor 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 Input3 31 2 72 3 43 1 43 21 2 72 3 40 0 

 

Sample Output-14題意：曹操在長江上建造了n個島嶼，這些島嶼之間要用橋串連起來，每座橋上都派兵駐守，現在周瑜手中有一個炸彈，他要派人去炸掉其中一座橋，他派去的人數必須大於這條橋上駐守的人數，問他最少要派多少人去注意：1、如果圖本來就不連通  不需要派人輸出02、如果橋上沒有人把守，派去一個人即可3、如果炸掉橋還不能斷開所有島嶼之間的的串連，輸出-1

#include<stdio.h>#include<string.h>#include<stack>#include<algorithm>#include<vector>#define MAX 1010#define MAXM 1000100#define INF 0x7ffffffusing namespace std;int n,m,mark;//mark記錄圖是否聯通 int head[MAX],ans;int low[MAX],dfn[MAX];int dfsclock,dcccnt;struct node{int beg,end,val,next;int cnt;//記錄橋是否存在 }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;//初始化為0表示沒有橋 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)//判斷重邊         {        flag=0;        continue;        }        if(!dfn[v])        {        tarjan(v,u);        low[u]=min(low[u],low[v]);        if(dfn[u]<low[v])//是橋    edge[i].cnt=edge[i^1].cnt=1; //標記這條邊是橋edge[i^1].cnt意思是這條邊的反向邊         }        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++)//遍曆所有點 {if(!dfn[i])//如果點沒被搜尋到證明這個圖不連通 {mark=0;return ;} }}void solve(){if(!mark)//圖不連通  不需要派人 直接輸出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);//尋找人數最少把守的橋 }if(ant==INF)//如果不能阻斷各島嶼之間的聯絡     ant=-1;if(ant==0)//如果橋上的人數為0 派一個人     ant=1;printf("%d\n",ant);}}int main(){while(scanf("%d%d",&n,&m),n|m){init();getmap();find();solve();}return 0;}

　　

hdoj 4738 Caocao's Bridges【雙連通分量求橋】