【POJ2152】Fire——树形DP

Fire

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 1294 Accepted: 669

Description

Country Z has N cities, which are numbered from 1 to N. Cities are connected by highways, and there is exact one path between two different cities. Recently country Z often caught fire, so the government decided to build some firehouses in some cities. Build a firehouse in city K cost W(K). W for different cities may be different. If there is not firehouse in city K, the distance between it and the nearest city which has a firehouse, can’t be more than D(K). D for different cities also may be different. To save money, the government wants you to calculate the minimum cost to build firehouses.

Input

The first line of input contains a single integer T representing the number of test cases. The following T blocks each represents a test case.

The first line of each block contains an integer N (1 < N <= 1000). The second line contains N numbers separated by one or more blanks. The I-th number means W(I) (0 < W(I) <= 10000). The third line contains N numbers separated by one or more blanks. The I-th number means D(I) (0 <= D(I) <= 10000). The following N-1 lines each contains three integers u, v, L (1 <= u, v <= N，0 < L <= 1000), which means there is a highway between city u and v of length L.

Output

For each test case output the minimum cost on a single line.

Sample Input

5
5
1 1 1 1 1
1 1 1 1 1
1 2 1
2 3 1
3 4 1
4 5 1
5
1 1 1 1 1
2 1 1 1 2
1 2 1
2 3 1
3 4 1
4 5 1
5
1 1 3 1 1
2 1 1 1 2
1 2 1
2 3 1
3 4 1
4 5 1
4
2 1 1 1
3 4 3 2
1 2 3
1 3 3
1 4 2
4
4 1 1 1
3 4 3 2
1 2 3
1 3 3
1 4 2

Sample Output

2
1
2
2
3

Source

POJ Monthly,Lou Tiancheng

Translation

给一棵带权树，每个节点需要被覆盖，覆盖节点i的方法为一下二者之一：

1. 用 wi 的价格在该点建消防站
2. 在该点距离 di 的范围内有至少一个消防站

求覆盖整棵树的最小代价

Solution

记dp[i][j]表示第i号节点被j号节点覆盖，且i号节点的子树被全部覆盖的最小代价，best[i]表示将i号节点的子树全部覆盖的最小代价，dis[i][j]表示树上两点i,j间的距离
dis[i][j]可以用 O(n2) 的时间预处理
dp[i][j]和best[i]可以这样转移

dp[i][j]=w[j]+∑k是i的儿子min(dp[k][j]−w[j],best[k])

best[i]=minnj=1dp[i][j]

其中，由于每一个dp[k][j]都包含了w[j]，所以我们在统计儿子时要将其去掉，并在最后加上
最后的答案便是best[root]，root可以任定

Code

/*
POJ2152
Author:Johann
*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define red(i, a, b) for(int i = (a); i >= (b); i--)
#define ll long long

inline int read() {
    char c = getchar(); int x = 0;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
    return x;
}

const int N = 1500;
const int inf = 1000000000;
struct edge{
    int from, to, len, nxt;
}e[N * 2];
int dp[N][N], dis[N][N];
int head[N], best[N], vis[N], w[N], d[N], lst[N];
int q[N];
int T, n, tail = 0, cnt = 0;

void addedge(int x, int y, int z) {
    e[++tail].from = x;
    e[tail].to = y;
    e[tail].len = z;
    e[tail].nxt = head[x];
    head[x] = tail;
}

void work(int root) {
    if (root == 1) lst[cnt = 1] = 1;
    memset(vis, 0, sizeof(vis));
    dis[root][root] = 0; 
    vis[root] = 1;
    int l = 1, r = 1;
    q[1] = root;
    while(l <= r) {
        int x = q[l]; l++;
        for(int i = head[x]; i != -1; i = e[i].nxt) {
            int v = e[i].to;
            if (!vis[v]) {
                vis[v] = 1; 
                if (root == 1) lst[++cnt] = v;
                dis[root][v] = dis[root][x] + e[i].len;
                q[++r] = v;
            }
        }
    }
}

void DP() {
    memset(vis, 0, sizeof(vis));
    red(k, n, 1) {
        int u = lst[k];
        vis[u] = 1;
        best[u] = inf;
        rep(j, 1, n) {
            if (dis[u][j] > d[u]) {
                dp[u][j] = inf;
                continue;
            }
            dp[u][j] = 0;
            for(int i = head[u]; i != -1; i = e[i].nxt) {
                int v = e[i].to;
                if (!vis[v]) continue;
                dp[u][j] += min(dp[v][j] - w[j], best[v]);
            }
            dp[u][j] += w[j];
            if (dp[u][j] < best[u]) best[u] = dp[u][j];
        }
    }
}

int main() {
    scanf("%d", &T);
    while(T--) {
        tail = 0;
        n = read();
        rep(i, 1, n) w[i] = read();
        rep(i, 1, n) d[i] = read();
        rep(i, 1, n) head[i] = -1;
        rep(i, 1, n - 1) {
            int x = read(), y = read(), z = read();
            addedge(x, y, z);
            addedge(y, x, z);
        }
        rep(i, 1, n) work(i);
        DP();
        printf("%d\n", best[1]);
    }
    return 0;
}

尾声

终于拿到新键盘了，写题的动力！
茶轴真棒！
千万不能把Codeforces的习惯带入POJ（可见BZOJ的优越），第一次体会到STL Queue被卡T的快感
陈启峰论文题……好吧的确不容易
老大在机房里开了监！控！
监！控！
重复一遍，HBH是心机婊

End.