[做题笔记] 2016/12/1

1000. 1004. dijkstra算法

1001. 递归，每一个点出发的最大路径=max（它所连点中出发路径的最大值+边权值）

1002. 广度优先搜索同时标记即可

1003. 样例只有一个。

【代码】

1000.

// Problem#: 19818// Submission#: 4937880// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University// Problem#: 19818// Submission#: 4937837// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <iostream>#include <memory.h>#define INF 1000000#define MAX 200using namespace std; int main(){    int arcs[MAX][MAX];    bool isvisit[MAX];    int min_weight[MAX];     int N;    int cases;     cin >> cases;     while (cases--)    {        cin >> N;        memset(isvisit, false, sizeof(isvisit));         for (int i = 0; i < N; i++)            for (int j = 0; j < N; j++)                cin >> arcs[i][j];         for (int i = 0; i < N; i++)            min_weight[i] = arcs[0][i];         isvisit[0] = true;        int _min;        int hold, result = 0, max_weight = 0;        for (int i = 1; i < N; i++)        {            _min = INF;                         for (int j = 0; j < N; j++)            {                if (!isvisit[j] && min_weight[j] < _min)                {                    _min = min_weight[j];                    hold = j;                 }            }               max_weight = max(max_weight, _min);            result += _min;             isvisit[hold] = true;             for (int j = 0; j < N; j++)                if (!isvisit[j] && arcs[hold][j] < min_weight[j])                    min_weight[j] = arcs[hold][j];        }         cout << max_weight << endl ;        if (cases != 0)            cout << endl;    }     return 0;}                                 

1001.

// Problem#: 19819// Submission#: 4937640// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include<iostream>#include<vector>#include<cstdlib>#include<cstdio>#include<memory.h>using namespace std;typedef struct edge{   int st;   int en;    int we;   edge( int a,int b,int c){st=a,en=b,we=c;}}e;vector<e> E[10010];bool vis[10010];int dfs(int k){   int i;   int max=0;   vis[k]=true;   for( i=0; i<E[k].size(); i++)   {           if( !vis[E[k][i].en]){          int temp=dfs(E[k][i].en);          if(E[k][i].we+temp>max )            max=E[k][i].we+temp;        }   }   vis[k]=0;   return max;}int main(){    int n,k,i,j,u,v,d;    while(cin>>n>>k)    {        memset(vis,0,sizeof(vis));        memset(E,0,sizeof(E));        for(i=0; i<n-1; i++)        {            cin>>u>>v>>d;      E[u].push_back(e(u,v,d));      E[v].push_back(e(v,u,d));        }       cout<< dfs(k)<<endl;    }}                                 

1002.

// Problem#: 19820// Submission#: 4937822// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University// Problem#: 4379  // Submission#: 1121129  // The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License  // URI: http://creativecommons.org/licenses/by-nc-sa/3.0/  // All Copyright reserved by Informatic Lab of Sun Yat-sen University  #include <cstdio>  #include <queue>  #include <cstring>  using namespace std;    int main()  {      int n, m;      int i, j;      queue<int> q;      int color[ 1001 ];      bool path[ 1001 ][ 1001 ];        memset( color, 0, sizeof( color ) );      memset( path, false, sizeof( path ) );        scanf( "%d%d", &n, &m );      while ( m-- ) {          scanf( "%d%d", &i, &j );          path[ i ][ j ] = true;          path[ j ][ i ] = true;      }        q.push( 1 );      color[ 1 ] = -1;      while ( !q.empty() ) {          i = q.front();          for ( j = 1; j <= n; j++ ) {              if ( path[ i ][ j ] ) {                  if ( color[ i ] == color[ j ] ) {                      printf( "no\n" );                      return 0;                  }                  if ( color[ j ] == 0 ) {                      q.push( j );                      color[ j ] = ( color[ i ] == 1 ? -1: 1 );                  }              }          }          q.pop();      }      printf( "yes\n" );        return 0;    }