poj 1192 树形dp

来源：互联网发布：原谅我红尘颠倒知乎编辑：程序博客网时间：2024/06/08 11:35

好久没刷过树形dp了，都快忘记了

dp【n】【0】表示不含有节点n时的最大权值和

dp【n】【1】表示含有节点n时的最大权值和

AC代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define MAX 0x3f3f3f3fstruct Edge{    int to, next;};struct Point{    int x, y;    int value;};int head[1100], tot;Edge edge[2200];Point point[1100];int N;int dp[1100][2];bool mark[1100];int add_adge( int a, int b ){    edge[tot].to = b;    edge[tot].next = head[a];    head[a] = tot++;    edge[tot].to = a;    edge[tot].next = head[b];    head[b] = tot++;    return 0;}int DFS( int pre, int pos ){    mark[pos] = true;    dp[pos][0] = 0;    for( int i = head[pos]; i != -1; i = edge[i].next ){        int to = edge[i].to;        if( to == pre ){            continue;        }        if( !mark[to] ){            DFS( pos, to );        }        dp[pos][0] = max( dp[pos][0], dp[to][0] );        dp[pos][0] = max( dp[pos][0], dp[to][1] );    }    dp[pos][1] = point[pos].value;    for( int i = head[pos]; i != -1; i = edge[i].next ){        int to = edge[i].to;        if( to == pre ){            continue;        }        if( !mark[to] ){            DFS( pos, to );        }        dp[pos][1] += max( 0, dp[to][1] );    }    return 0;}int main(){    while( scanf( "%d", &N ) != EOF ){        tot = 0;        memset( head, -1, sizeof( head ) );        for( int i = 0; i < N; i++ ){            cin >> point[i].x >> point[i].y >> point[i].value;        }        for( int i = 0; i < N; i++ ){            for( int j = i + 1; j < N; j++ ){                if( abs( point[i].x - point[j].x ) + abs( point[i].y - point[j].y ) == 1 ){                    add_adge( i, j );                }            }        }        for( int i = 0; i < N; i++ ){            for( int j = 0; j < 2; j++ ){                dp[i][j] = -MAX;            }        }        memset( mark, false, sizeof( mark ) );        DFS( -1, 0 );        int ans = -MAX;        for( int i = 0; i < N; i++ ){            ans = max( ans, dp[i][1] );            ans = max( ans, dp[i][0] );        }        cout << ans << endl;    }    return 0;}

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