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BestCoder Round #1 1001 && 1002 hdu 4857 4858

hdu 4857 逃生

第一题是拓扑排序,不是按照字典序最小输出,而是要使较小的数排在最前面。。赛后弄了好久,才比较明白,我一直以为 反向建图,i从1到n,开始深搜dfs( i ),对i点的边,由小到大继续搜一下,同时标记搜过的数,搜过之后就不再搜,搜到底之后ans[cnt++] = u;这样顺序输出就是答案,后来经过超哥指点,才明白深搜贪心是错的。只有 反向建图,用优先队列把较大的数尽量排在前面,然后反序输出才是正解。。

  1 #include<iostream>  2 #include<cstring>  3 #include<algorithm>  4 #include<cstdio>  5 #include<string>  6 #include<queue>  7 #include<cmath>  8 #include<vector>  9  10 using namespace std; 11  12 #define mnx 104000 13 #define ll long long 14 #define inf 0x3f3f3f3f 15 #define lson l, m, rt << 1 16 #define rson m+1, r, rt << 1 | 1 17  18 vector<int> vet[mnx]; 19 int cnt, ans[mnx], vis[mnx]; 20 int main(){ 21     int cas; 22     scanf( "%d", &cas ); 23     while( cas-- ){ 24         for( int i = 0; i < mnx; i++ ){ 25             vet[i].clear(); 26         } 27         memset( vis, 0, sizeof(vis) ); 28         cnt = 0; 29         int n, m; 30         scanf( "%d%d", &n, &m ); 31         for( int i = 0; i < m; i++ ){ 32             int u, v; 33             scanf( "%d%d", &u, &v ); 34             vis[u]++; 35             vet[v].push_back( u ); 36         } 37         priority_queue<int> que; 38         for( int i = 1; i <= n; i++ ){ 39             if( !vis[i] ) que.push( i ); 40         } 41         while( !que.empty() ){ 42             int u = que.top(); que.pop(); 43             for( int i = 0; i < vet[u].size(); i++ ){ 44                 vis[vet[u][i]]--; 45                 if( vis[vet[u][i]] == 0 ){ 46                     que.push( vet[u][i] ); 47                 } 48             } 49             ans[cnt++] = u; 50         } 51         for( int i = cnt-1; i >= 0; i-- ){ 52             if( i == 0 ){ 53                 cout<<ans[i]<<endl; 54             } 55             else cout<<ans[i]<<" "; 56         } 57     } 58     return 0; 59 } 60 /*给出几组数据大家试一下  61 10 62 3 1 63 3 1 64 answer : 3 1 2 65 3 1 66 1 3 67 answer : 1 2 3 68 4 3 69 3 1 70 4 1 71 2 4 72 answer : 2 3 1 4 73 9 12 74 1 2 75 2 8 76 2 9 77 2 4 78 4 3 79 4 5 80 4 7 81 9 3 82 9 6 83 9 5 84 8 5 85 8 7 86 answer : 1 2 4 9 3 8 5 6 7 87 按照字典序最小输出  88 int vv[mnx], first[mnx], nxt[mnx], e, d[mnx], ans[mnx], n; 89 void add( int u, int v ){ 90     vv[e] = v, nxt[e] = first[u], first[u] = e++; 91 } 92 int main(){ 93     n = 8; 94     memset( d, 0, sizeof(d) ); 95     memset( first, -1, sizeof(first) ); 96     for( int i = 0; i < 6; i++ ){ 97         int u, v; 98         scanf( "%d%d", &u, &v ); 99         add( u, v );100         d[v]++;101     }102     int cnt = 0;103     priority_queue< int, vector<int>, greater<int> > q;104     for( int i = 1; i <= n; i++ ){105         if( d[i] == 0 ) q.push( i );106     }107     while( !q.empty() ){108         int u = q.top(); q.pop();109         ans[cnt++] = u;110         for( int i = first[u]; i != -1; i = nxt[i] ){111             int v = vv[i];112             d[v]--;113             if( d[v] == 0 ) q.push( v );114         }115         for( int i = 0; i < n; i++ ){116             cout<<ans[i]<<" ";117         }118         cout<<endl;119     }120     return 0;121 }122 123 */
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hdu 4858 项目管理

第二题数据比较水,暴力可以就过了。。赛后问了超哥正解,好像是图的分块:按点的度数和sqrtn分块,度数大于sqrtn的点最多有sqrtn个,把大点和大点建图,更新首先直接更新自己的值,然后更新小点的时候,一并更新小点连接的所有大点的答案,更新大点的时候,把相邻的大点也更新答案,询问的话,小点直接for循环累加val值,大点直接返回答案。。

 1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 #include<cstdio> 5 #include<string> 6 #include<queue> 7 #include<cmath> 8 #include<vector> 9 10 using namespace std;11 12 #define mnx 10400013 #define ll long long14 #define inf 0x3f3f3f3f15 #define lson l, m, rt << 116 #define rson m+1, r, rt << 1 | 117 18 vector<int> g1[mnx], g2[mnx];19 int d[mnx], val[mnx], ans[mnx], n, m;;20 void init(){21     for( int i = 1; i <= n; i++ ){22         if( d[i] > 333 )23             for( int j = 0; j < g1[i].size(); j++ )24                 if( d[g1[i][j]] > 333 ) 25                     g2[i].push_back( g1[i][j] );26     }27 }28 void upd(){29     int u, v;30     scanf( "%d%d", &u, &v );31     val[u] += v;32     if( ans[u] <= 333 ){33         for( int i = 0; i < g1[u].size(); i++ ){34             int V = g1[u][i];35             if( d[V] > 333 ) ans[V] += v;36         }37     }38     else for( int i = 0; i < g2[u].size(); i++ ){39         ans[ g2[u][i] ] += v;40     }41 }42 void cal(){43     int u;44     scanf( "%d", &u );45     if( d[u] > 333 ) printf( "%d\n", ans[u] );46     else{47         int sol = 0;48         for( int i = 0; i < g1[u].size(); i++ ){49             sol += val[ g1[u][i] ];50         }51         printf( "%d\n", sol );52     }53 }54 int main(){55     int cas;56     scanf( "%d", &cas );57     while( cas-- ){58         for( int i = 0; i < mnx; i++ ){59             g1[i].clear(), g2[i].clear();60         }61         memset( ans, 0, sizeof(ans) );62         memset( d, 0, sizeof(d) );63         memset( val, 0, sizeof(val) );64         scanf( "%d%d", &n, &m );65         for( int i = 0; i < m; i++ ){66             int u, v;67             scanf( "%d%d", &u, &v );68             g1[u].push_back( v ), g1[v].push_back( u );69             d[u]++, d[v]++;70         }71         init();72         int q;73         scanf( "%d", &q );74         while( q-- ){75             int cmd;76             scanf( "%d", &cmd );77             if( cmd == 0 ){78                 upd();79             }80             else cal();81         }82     }83     return 0;84 }
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