uva_10859_Placing Lampposts( 樹形DP )

題意：在一個n個節點，m條邊的無向無環圖中的節點中放置街燈，一個結點放置街燈那麼與這個結點爲頂點的邊都被照亮，使得每條邊都有被照亮的最少放置街燈的前提下，使得被兩個街燈照亮的邊儘量的多分析：在這裏的被兩個燈照亮的邊儘量的多，變一下：使得被一個燈照亮的邊儘量少無向無環圖也就是森林, 對於求兩個變量的最小設一個等式: x = aM+c, 其中a是需要最少的燈，c是一個燈照亮的邊數量，目標就是使得x最小，因爲a = x/M, c = x%M狀態：d[i,j]表示以i爲root的子樹，i的parent街燈放置的情況 x的數值狀態轉移看Code比較清晰，不好敘述Code:#include <set>#include <map>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <deque>#include <vector>#include <cstdio>#include <bitset>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;#define DIR     4#define DIM     2#define STATUS  2#define MAXM    1000 + 10#define MAXN    1000 + 10#define oo      (~0u)>>1#define INF     0x3F3F3F3F#define REPI(i, s, e)        for(int i = s; i <= e; i ++)#define REPD(i, e, s)         for(int i = e; i >= s; i --)static const double EPS = 1e-5;vector<int> vertex[MAXN];int f[MAXN][STATUS], visit[MAXN][STATUS], M;inline int dp(int x, int y, int p){        if( visit[x][y] ) {                return f[x][y];        }        visit[x][y] = 1;        // x put a light        f[x][y] = M;        for(int i = 0; i < vertex[x].size(); i ++) {                int k = vertex[x][i];                if( k == p ) {                        continue;                }                f[x][y] += dp(k, 1, x);        }        if( p >= 0 && !y ) {                f[x][y] += 1;        }        // x not put a light        if( y || p < 0 ) {                int tmp = 0;                for(int i = 0; i < vertex[x].size(); i ++) {                        int k = vertex[x][i];                        if( k == p ) {                                continue;                        }                        tmp += dp(k, 0, x);                }                if( p >= 0 ) {                        tmp += 1;                }                f[x][y] = min(f[x][y], tmp);        }        return f[x][y];}int main(int argc, char const *argv[]){#ifndef ONLINE_JUDGE        freopen("test.in", "r", stdin);#endif        int cas, n, m, u, v;        M = 2000;        scanf("%d", &cas);        REPI(k, 1, cas) {                scanf("%d %d", &n, &m);                REPI(i, 0, n) {                        memset(visit[i], 0, sizeof(visit[i]));                        vertex[i].erase(vertex[i].begin(), vertex[i].end());                }                REPI(i, 1, m) {                        scanf("%d %d", &u, &v);                        vertex[u].push_back(v);                        vertex[v].push_back(u);                }                int ans = 0;                REPI(i, 0, n-1) {                        if( visit[i][1] ) {                                continue;                        }                        ans += dp(i, 1, -1);                }                printf("%d %d %d\n", ans/M, m-ans%M, ans%M);        }        return 0;}