题目链接

问题分析

其实是比较明显的动态DP。

懒于再推一遍式子,直接用 最小权点覆盖=全集-最大权独立集,然后就和这道题一样了。题解可以看这里

然后必须选或者不选的话,就直接把相应的点权变成\(-\infty\)或\(\infty\)就好了。如果是必须选,最后答案里不要忘了加回原来的值。

参考程序

#include <bits/stdc++.h>
using namespace std; const long long Maxn = 100010;
const long long INF = 10000000010;
struct matrix {
long long A[ 2 ][ 2 ];
matrix() {
A[ 0 ][ 0 ] = A[ 0 ][ 1 ] = A[ 1 ][ 0 ] = A[ 1 ][ 1 ] = -INF;
return;
}
matrix( long long *T ) {
A[ 0 ][ 0 ] = A[ 0 ][ 1 ] = T[ 0 ];
A[ 1 ][ 0 ] = T[ 1 ];
A[ 1 ][ 1 ] = -INF;
return;
}
matrix( long long a, long long b, long long c, long long d ) {
A[ 0 ][ 0 ] = a; A[ 0 ][ 1 ] = b; A[ 1 ][ 0 ] = c; A[ 1 ][ 1 ] = d;
return;
}
inline matrix operator * ( const matrix Other ) const {
return matrix( max( A[ 0 ][ 0 ] + Other.A[ 0 ][ 0 ], A[ 0 ][ 1 ] + Other.A[ 1 ][ 0 ] ),
max( A[ 0 ][ 0 ] + Other.A[ 0 ][ 1 ], A[ 0 ][ 1 ] + Other.A[ 1 ][ 1 ] ),
max( A[ 1 ][ 0 ] + Other.A[ 0 ][ 0 ], A[ 1 ][ 1 ] + Other.A[ 1 ][ 0 ] ),
max( A[ 1 ][ 0 ] + Other.A[ 0 ][ 1 ], A[ 1 ][ 1 ] + Other.A[ 1 ][ 1 ] ) );
}
};
struct edge {
long long To, Next;
edge() : To( 0 ), Next( 0 ) {}
edge( long long _To, long long _Next ) : To( _To ), Next( _Next ) {}
};
edge Edge[ Maxn << 1 ];
long long Start[ Maxn ], UsedEdge;
inline void AddEdge( long long x, long long y ) {
Edge[ ++UsedEdge ] = ( edge ) { y, Start[ x ] };
Start[ x ] = UsedEdge;
return;
}
long long n, m, Cost[ Maxn ], NowCost[ Maxn ], Sum;
char Ch[ 10 ];
long long Deep[ Maxn ], Father[ Maxn ], Size[ Maxn ], Son[ Maxn ], Top[ Maxn ], Index[ Maxn ], Ref[ Maxn ], Dfn[ Maxn ], Used;
long long Dp[ Maxn ][ 2 ], LDp[ Maxn ][ 2 ];
matrix Tree[ Maxn << 2 ]; void Build_Cut();
void Build();
void Change( long long u, long long Key );
matrix Query( long long Ind, long long Left, long long Right, long long L, long long R ); int main() {
scanf( "%lld%lld", &n, &m ); scanf( "%s", Ch );
for( long long i = 1; i <= n; ++i ) scanf( "%lld", &Cost[ i ] );
for( long long i = 1; i < n; ++i ) {
long long x, y; scanf( "%lld%lld", &x, &y );
AddEdge( x, y ); AddEdge( y, x );
}
Build_Cut();
Build();
for( long long i = 1; i <= m; ++i ) {
long long a, x, b, y;
scanf( "%lld%lld%lld%lld", &a, &x, &b, &y);
if( x == 0 ) { Sum -= Cost[ a ]; Sum += INF; Change( a, INF ); }
else { Sum -= Cost[ a ]; Sum += -INF; Change( a, -INF ); }
if( y == 0 ) { Sum -= Cost[ b ]; Sum += INF; Change( b, INF ); }
else { Sum -= Cost[ b ]; Sum += -INF; Change( b, -INF ); } matrix Temp = Query( 1, 1, n, Index[ 1 ], Dfn[ 1 ] );
long long Ans = max( Temp.A[ 0 ][ 0 ], Temp.A[ 1 ][ 0 ] );
Ans = Sum - Ans;
if( x == 1 ) Ans = Ans + INF + Cost[ a ];
if( y == 1 ) Ans = Ans + INF + Cost[ b ];
if( Ans >= INF ) printf( "-1\n" ); else printf( "%lld\n", Ans ); if( x == 0 ) { Sum -= INF; Sum += Cost[ a ]; Change( a, Cost[ a ] ); }
else { Sum -= -INF; Sum += Cost[ a ]; Change( a, Cost[ a ] ); }
if( y == 0 ) { Sum -= INF; Sum += Cost[ b ]; Change( b, Cost[ b ] ); }
else { Sum -= -INF; Sum += Cost[ b ]; Change( b, Cost[ b ] ); }
}
return 0;
} void Dfs1( long long u, long long Fa ) {
Deep[ u ] = Deep[ Fa ] + 1;
Father[ u ] = Fa;
Size[ u ] = 1;
for( long long t = Start[ u ]; t; t = Edge[ t ].Next ) {
long long v = Edge[ t ].To;
if( v == Fa ) continue;
Dfs1( v, u );
if( Size[ v ] > Size[ Son[ u ] ] ) Son[ u ] = v;
Size[ u ] += Size[ v ];
}
return;
} void Dfs2( long long u, long long Fa ) {
if( Son[ u ] ) {
Top[ Son[ u ] ] = Top[ u ];
Index[ Son[ u ] ] = ++Used;
Ref[ Used ] = Son[ u ];
Dfs2( Son[ u ], u );
}
for( long long t = Start[ u ]; t; t = Edge[ t ].Next ) {
long long v = Edge[ t ].To;
if( v == Fa || v == Son[ u ] ) continue;
Top[ v ] = v; Index[ v ] = ++Used; Ref[ Used ] = v;
Dfs2( v, u );
}
return;
} void Build_Cut() {
Dfs1( 1, 0 );
Top[ 1 ] = 1; Index[ 1 ] = ++Used; Ref[ Used ] = 1;
Dfs2( 1, 0 );
for( int i = 1; i <= n; ++i ) Dfn[ Top[ i ] ] = max( Dfn[ Top[ i ] ], Index[ i ] );
return;
} void Dfs3( long long u, long long Fa ) {
LDp[ u ][ 1 ] = NowCost[ u ];
for( long long t = Start[ u ]; t; t = Edge[ t ].Next ) {
long long v = Edge[ t ].To;
if( v == Fa || v == Son[ u ] ) continue;
Dfs3( v, u );
LDp[ u ][ 0 ] += max( Dp[ v ][ 0 ], Dp[ v ][ 1 ] );
LDp[ u ][ 1 ] += Dp[ v ][ 0 ];
}
if( Son[ u ] ) Dfs3( Son[ u ], u );
Dp[ u ][ 0 ] = LDp[ u ][ 0 ] + max( Dp[ Son[ u ] ][ 0 ], Dp[ Son[ u ] ][ 1 ] );
Dp[ u ][ 1 ] = LDp[ u ][ 1 ] + Dp[ Son[ u ] ][ 0 ];
return;
} void RealBuild( long long Ind, long long Left, long long Right ) {
if( Left == Right ) {
Tree[ Ind ] = matrix( LDp[ Ref[ Left ] ] );
return;
}
long long Mid = ( Left + Right ) >> 1;
RealBuild( Ind << 1, Left, Mid );
RealBuild( Ind << 1 | 1, Mid + 1, Right );
Tree[ Ind ] = Tree[ Ind << 1 ] * Tree[ Ind << 1 | 1 ];
return;
} matrix Query( long long Ind, long long Left, long long Right, long long L, long long R ) {
if( L <= Left && Right <= R ) return Tree[ Ind ];
long long Mid = ( Left + Right ) >> 1;
if( R <= Mid ) return Query( Ind << 1, Left, Mid, L, R );
if( L > Mid ) return Query( Ind << 1 | 1, Mid + 1, Right, L, R );
return Query( Ind << 1, Left, Mid, L, R ) * Query( Ind << 1 | 1, Mid + 1, Right, L, R );
} void Build() {
for( long long i = 1; i <= n; ++i ) NowCost[ i ] = Cost[ i ];
for( long long i = 1; i <= n; ++i ) Sum += NowCost[ i ];
Dfs3( 1, 0 );
RealBuild( 1, 1, n );
return;
} void Update( long long Ind, long long Left, long long Right, long long x ) {
if( Left == Right ) {
Tree[ Ind ] = matrix( LDp[ Ref[ Left ] ] );
return;
}
long long Mid = ( Left + Right ) >> 1;
if( x <= Mid ) Update( Ind << 1, Left, Mid, x );
if( x > Mid ) Update( Ind << 1 | 1, Mid + 1, Right, x );
Tree[ Ind ] = Tree[ Ind << 1 ] * Tree[ Ind << 1 | 1 ];
return;
} void Change( long long u, long long Key ) {
matrix Last = Query( 1, 1, n, Index[ Top[ u ] ], Dfn[ Top[ u ] ] );
LDp[ u ][ 1 ] += -NowCost[ u ] + Key; NowCost[ u ] = Key;
Update( 1, 1, n, Index[ u ] );
matrix Temp = Query( 1, 1, n, Index[ Top[ u ] ], Dfn[ Top[ u ] ] );
u = Father[ Top[ u ] ];
while( u ) {
matrix TTT = Query( 1, 1, n, Index[ Top[ u ] ], Dfn[ Top[ u ] ] );
LDp[ u ][ 0 ] += -max( Last.A[ 0 ][ 0 ], Last.A[ 1 ][ 0 ] ) + max( Temp.A[ 0 ][ 0 ], Temp.A[ 1 ][ 0 ] );
LDp[ u ][ 1 ] += -Last.A[ 0 ][ 0 ] + Temp.A[ 0 ][ 0 ];
Last = TTT;
Update( 1, 1, n, Index[ u ] );
Temp = Query( 1, 1, n, Index[ Top[ u ] ], Dfn[ Top[ u ] ] );
u = Father[ Top[ u ] ];
}
return;
}

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