#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[1505][2];
int vis[1505], edges[1505];
int len,root;
struct node {
int now, next;
} tree[3005];//因为要双向建,所以要开2倍大小
void add(int x, int y) { //建树,单向建 双向建都可AC
tree[len].now = y;
tree[len].next = edges[x];//头插法
edges[x] = len;//记录节点x在边集中的编号
len++;
tree[len].now = x;
tree[len].next = edges[y];//头插法
edges[y] = len;//记录节点y在边集中的编号
len++;
}
void dfs(int root) {
vis[root] = 1;
dp[root][1] = 1;
dp[root][0] = 0;
for(int i = edges[root]; i != -1; i = tree[i].next) {
int k = tree[i].now;
if(!vis[k]) {
dfs(k);
dp[root][0] += dp[k][1];//父亲不放
dp[root][1] += min(dp[k][1], dp[k][0]);//父亲放
}
}
}
int main() {
freopen("data.in", "r", stdin);
int t, x, y, n, i, j;
while(scanf("%d", &t) == 1) {
root = -1;
len = 0;
memset(dp, 0, sizeof(dp));
memset(vis, 0, sizeof(vis));
memset(edges, -1, sizeof(edges));
for(int j = 1; j <= t; ++j) {
scanf("%d:(%d)", &x, &n);
if(root == -1)
root = x;
for(int i = 1; i <= n; ++i) {
scanf("%d", &y);
add(x, y);
}
}
dfs(root);
printf("%d\n",min(dp[root][0],dp[root][1]));
}
return 0;
}
HDU1054 && POJ1463:Strategic game(树形DP)
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转载自blog.csdn.net/ccshijtgc/article/details/81042677
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