2017年 10 月 9 日 機房模擬賽

T1是揹包問題，但是我不知道正解（正解是狀態巧妙的揹包問題）

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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裸狀壓，然而我發現我根本不會狀壓，這道題我們先把所有單層合法狀態枚舉出來，然後再進行每一層的DFS，然後統計次數
dfs過程中判斷掉錯誤情況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是單調棧，我們維護幾個數組分別表示從這一行中每個點最多能到的左邊的位置從這一行中這個點最多能到的右邊的位置，過程中我們維護單調棧即可，哇好難啊我根本不會啊，今天好多人AK了，而且是考了一個小時就已經AK之後無所事事了，還寫了好多種其他方法的AK大佬，永遠不可及了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是左邊的棧,R是右邊的棧
        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這一行 
        } 
        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直接拓撲排序然後計算一下，注意有負數，然後還有一種方法是把超級源連向所有入度爲0的點，把所有出度爲0的點連向超級匯，然後跑一邊最短路即可

#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;
}