2017年 10 月 9 日 机房模拟赛

T1是背包问题，但是我不知道正解（正解是状态巧妙的背包问题）

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#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 2000 + 100;int N, M, dp[ MAXN + 10 ], tail;struct P { int w, a, b; } p[ MAXN + 10 ];bool cmp( const P &A, const P &B ) { return A.w < B.w; }int main( ) {    freopen( "birthday.in", "r", stdin );    freopen( "birthday.out", "w", stdout );    scanf( "%d%d", &N, &M );     for( register int i = 1; i <= N; i++ )        scanf( "%d%d%d", &p[i].w, &p[i].a, &p[i].b );    for( register int i = 1; i <= N; i++ ) {        for( register int j = M; j >= 1; j-- ) {            if( p[i].w > j ) break;            int up = j / p[i].w;            int tmp = 0;            for( register int k = 1; k <= up; k++ )                 tmp = max( tmp, dp[ j - p[i].w * k ] + p[i].a * k );             dp[j] = max( dp[j], tmp + p[i].b );         }    }    printf( "%d\n", dp[M] );    return 0;}

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T2裸状压，然而我发现我根本不会状压，这道题我们先把所有单层合法状态枚举出来，然后再进行每一层的DFS，然后统计次数
dfs过程中判断掉错误情况continue掉即可

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#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;const int mod = 100000000 ;int Lim, N, M, mp[15][15], lim[15], dp[15][ 1 << 14 ], situ[ 1 << 14 ], top;int Dfs( int pos, int t ) {    if( pos == N + 1 ) return 1;    if( dp[pos][t] != -1 ) return dp[pos][t];    int tmp = 0;    for( register int i = 1; i <= top; i++ ) {        if( situ[i] & lim[pos] ) continue;        if( situ[i] & t )        continue;        tmp = ( tmp + Dfs( pos + 1, situ[i] ) ) % mod;    }    return dp[pos][t] = tmp;}void dfs( int pos, int t ) {    if( pos == M + 1 ) { situ[++top] = t; return; }    if( pos == 1 || ( t & ( 1 << ( pos - 2 ) ) ) == 0 )        dfs( pos + 1, t + ( 1 << ( pos - 1 ) ) );    dfs( pos + 1, t );}int main( ) {    freopen( "chess.in", "r", stdin );    freopen( "chess.out", "w", stdout );    scanf( "%d%d", &N, &M );    Lim = ( 1 << M ) - 1; dfs( 1, 0 );    for( register int i = 1; i <= N; i++ )         for( register int j = 1; j <= M; j++ ) {            scanf( "%d", &mp[i][j] );            if( !mp[i][j] ) lim[i] |= ( 1 << ( j - 1 ) );        }    memset( dp, -1, sizeof(dp) );    dp[0][0] = 1;    printf( "%d", Dfs( 1, 0 ) );    return 0;}

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T3是单调栈，我们维护几个数组分别表示从这一行中每个点最多能到的左边的位置从这一行中这个点最多能到的右边的位置，过程中我们维护单调栈即可，哇好难啊我根本不会啊，今天好多人AK了，而且是考了一个小时就已经AK之后无所事事了，还写了好多种其他方法的AK大佬，永远不可及了w…

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;int up[1005][1005] , mp[1005][1005] ;int N , M , SL[1005] , SR[1005] , maxn , L[1005] , R[1005] , top ;void solve( ) {    for( register int i = 1; i <= N; i++ ) {        memset( L, 0, sizeof( L ) );//L是左边的栈,R是右边的栈        top = 0;        for( register int j = 1 ; j <= M; j++ ) {            while( top && up[i][ L[top] ] >= up[i][j] ) top--;            SL[j] = L[top] + 1; L[ ++top ] = j;//SL表示i这一行         }         memset( R, 0, sizeof( R ) ); top = 0; R[0] = M + 1;        for( register int j = M; j; j-- ) {            while( top && up[i][ R[top] ] >= up[i][j] ) top--;            SR[j] = R[top] - 1; R[++top] = j;        }         for( register int j = 1; j <= M; j++ )             maxn = max( maxn, ( SR[j] - SL[j] + 1 ) * up[i][j] );    }    printf( "%d", maxn );}int main( ) {    freopen( "question.in", "r", stdin );    freopen( "question.out", "w", stdout );    scanf( "%d%d", &N, &M );    for( register int i = 1; i <= N; i++ )         for( register int j = 1; j <= M; j++ ) {            scanf( "%d", &mp[i][j] );            if( mp[i][j]) up[i][j] = up[ i - 1 ][j] + 1;        }    solve();    return 0;}

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T4直接拓扑排序然后计算一下，注意有负数，然后还有一种方法是把超级源连向所有入度为0的点，把所有出度为0的点连向超级汇，然后跑一边最短路即可

#include <cstdio>#include <cstring>#include <iostream>#include <queue>using namespace std;const int MAXN = 2000000 + 10;int head[MAXN], tail, in[MAXN], out[MAXN], n, m;long long ans[MAXN], a[MAXN];struct Line{ int to, nxt; }line[ MAXN * 2 ];void add_line( int from, int to ) {    line[++tail].nxt = head[from];    line[tail].to = to;    head[from] = tail;    out[from]++; in[to]++;}void solve( ) {    long long ass = -214748123LL;    for( register int i = 1; i <= n; i++ ) ans[i] = -214748123LL;    queue<int>q; while( !q.empty() ) q.pop();    for( register int i = 1; i <= n; i++ )         if( in[i] == 0 ) {            q.push( i );            ans[i] = a[i];        }    while( !q.empty() ){        int u = q.front();  q.pop();        for( register int i = head[u]; i; i = line[i].nxt ) {            int v = line[i].to;            ans[v] = max( ans[v], ans[u] + a[v] );            in[v]--;            if( in[v] == 0 ) q.push(v);        }    }    for( register int i = 1; i <= n; i++ ) if( out[i] == 0 ) ass = max( ass, ans[i] );    printf( "%I64d\n", ass );}int main( ) {    freopen( "road.in", "r", stdin );    freopen( "road.out", "w", stdout );    scanf( "%d%d", &n, &m );    for( register int i = 1; i <= n; i++ ) scanf( "%I64d", &a[i] );    for( register int i = 1; i <= m; i++ ) { int ff, tt;        scanf( "%d%d", &ff, &tt );        add_line( ff, tt );    }    solve();    return 0;}