bzoj 3611
和BZOJ消耗站一样，先将那个询问的简图构建出来，然后就是简单的树形DP。
（倍增数组开小了，然后就狂WA，自己生成的极限数据深度又没有那么高，链又奇迹般正确）
1 #include <cstdio> 2 #include <cstring> 3 #include <vector> 4 #include <algorithm> 5 #define min(a,b) ((a)<(b)?(a):(b)) 6 #define max(a,b) ((a)>(b)?(a):(b)) 7 #define oo 0x3f3f3f3f3f3f3f3f 8 #define N 1000010 9 #define P 19 10 using namespace std; 11 12 typedef long long dnt; 13 14 int n, m; 15 vector<int> g[N]; 16 int dfn[N], anc[N][P+1], dep[N], idc; 17 int head[N], ikey[N], dest[N+N], next[N+N], etot; 18 dnt dp[N][4]; 19 int qcnt, aa[N], stk[N], top; 20 dnt ans[3]; 21 22 void adde( int u, int v ) { 23 etot++; 24 next[etot] = head[u]; 25 dest[etot] = v; 26 head[u] = etot; 27 } 28 void dfs( int u ) { 29 dfn[u] = ++idc; 30 for( int p=1; p<=P; p++ ) 31 anc[u][p] = anc[anc[u][p-1]][p-1]; 32 for( int t=0; t<g[u].size(); t++ ) { 33 int v=g[u][t]; 34 if( v==anc[u][0] ) continue; 35 anc[v][0] = u; 36 dep[v] = dep[u]+1; 37 dfs(v); 38 } 39 } 40 bool cmp( int u, int v ) { 41 return dfn[u]<dfn[v]; 42 } 43 int lca( int u, int v ) { 44 if( dep[u]<dep[v] ) swap(u,v); 45 int t=dep[u]-dep[v]; 46 for( int p=0; t; t>>=1,p++ ) 47 if( t&1 ) u=anc[u][p]; 48 if( u==v ) return u; 49 for( int p=P; p>=0 && anc[u][0]!=anc[v][0]; p-- ) 50 if( anc[u][p]!=anc[v][p] ) u=anc[u][p], v=anc[v][p]; 51 return anc[u][0]; 52 } 53 void build() { 54 stk[top=1] = 1; 55 head[1] = 0; 56 57 sort( aa+1, aa+1+qcnt, cmp ); 58 59 etot = 0; 60 for( int i=2-(aa[1]!=1); i<=qcnt; i++ ) { 61 int ca=lca(aa[i],stk[top]); 62 while( dep[ca]<dep[stk[top]] ) { 63 int u; 64 u = stk[top--]; 65 if( dep[ca]>=dep[stk[top]] ) { 66 if( ca!=stk[top] ) { 67 stk[++top] = ca; 68 head[ca] = 0; 69 } 70 adde( stk[top], u ); 71 break; 72 } 73 adde( stk[top], u ); 74 } 75 stk[++top] = aa[i]; 76 head[aa[i]] = 0; 77 } 78 for( int i=top; i>=2; i-- ) 79 adde( stk[i-1], stk[i] ); 80 } 81 dnt sml, big, sum, cnt; 82 void dodp( int u ) { 83 if( ikey[u] ) { 84 dp[u][0] = 0; 85 dp[u][1] = 0; 86 dp[u][2] = 1; 87 dp[u][3] = 0; 88 } else { 89 dp[u][0] = oo; 90 dp[u][1] = -oo; 91 dp[u][2] = 0; 92 dp[u][3] = 0; 93 } 94 for( int t=head[u]; t; t=next[t] ) { 95 int v=dest[t]; 96 dnt w=dep[v]-dep[u]; 97 dodp(v); 98 dp[u][0] = min( dp[u][0], dp[v][0]+w ); 99 dp[u][1] = max( dp[u][1], dp[v][1]+w ); 100 dp[u][2] += dp[v][2]; 101 dp[u][3] += dp[v][3]+w*dp[v][2]; 102 } 103 if( ikey[u] ) { 104 sml = 0; 105 big = 0; 106 sum = 0; 107 cnt = 1; 108 } else { 109 sml = oo; 110 big = -oo; 111 sum = 0; 112 cnt = 0; 113 } 114 for( int t=head[u]; t; t=next[t] ) { 115 int v=dest[t]; 116 dnt w=dep[v]-dep[u]; 117 ans[0] = min( ans[0], sml+w+dp[v][0] ); 118 ans[1] = max( ans[1], big+w+dp[v][1] ); 119 ans[2] += cnt*(dp[v][3]+w*dp[v][2]) + sum*dp[v][2]; 120 sml = min( sml, dp[v][0]+w ); 121 big = max( big, dp[v][1]+w ); 122 cnt += dp[v][2]; 123 sum += dp[v][3]+w*dp[v][2]; 124 } 125 } 126 int main() { 127 scanf( "%d", &n ); 128 for( int i=1,u,v; i<n; i++ ) { 129 scanf( "%d%d", &u, &v ); 130 g[u].push_back( v ); 131 g[v].push_back( u ); 132 } 133 anc[1][0] = 1; 134 dep[1] = 0; 135 dfs(1); 136 scanf( "%d", &m ); 137 for( int i=1; i<=m; i++ ) { 138 scanf( "%d", &qcnt ); 139 for( int j=1; j<=qcnt; j++ ) 140 scanf( "%d", aa+j ); 141 142 if( qcnt==1 ) { 143 printf( "0 0 0\n" ); 144 continue; 145 } 146 147 build(); 148 149 for( int j=1; j<=qcnt; j++ ) 150 ikey[aa[j]] = 1; 151 ans[0] = oo; 152 ans[1] = 0; 153 ans[2] = 0; 154 dodp(1); 155 for( int j=1; j<=qcnt; j++ ) 156 ikey[aa[j]] = 0; 157 158 printf( "%lld %lld %lld\n", ans[2], ans[0], ans[1] ); 159 } 160 }
