牛客练习赛76 D 魔物消灭计划 (斯坦纳树)
作者:互联网
题目链接:https://ac.nowcoder.com/acm/contest/10845/D
斯坦纳树:联通若干特殊点的最小生成树
该题中因为同种宝石间传送无需花费代价,所以把同种宝石的城市缩成一个特殊点,然后求斯坦纳树即可
斯坦纳树求解:
设 \(dp[i][S]\) 表示以 \(i\) 为根,联通特殊点的状态为 \(S\) 花费的最小代价
转移分为两部分
- \(dp[i][S] = min(dp[i][S], dp[i][sub] + dp[i][S ^ sub]\)
其中枚举子集使用以下技巧 (复杂度为 $O(3^n)):
for(int sub = S & (S - 1) ; sub ; sub = S & (sub - 1))
- \(dp[i][S] = min(dp[i][S], dp[j][S] + w[i][j])\)
第二部分转移就是三角形不等式,dijstra即可
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
#define mk make_pair
const int maxn = 610;
int n, m, k, x, y;
int a[maxn], id[maxn];
int dp[maxn][(1 << 10) + 10];
int h[maxn], cnt;
struct E{
int to, next, cost;
}e[maxn << 1];
void add(int u, int v, int w){
e[++cnt].to = v;
e[cnt].cost = w;
e[cnt].next = h[u];
h[u] = cnt;
}
priority_queue<P, vector<P>, greater<P> > q;
int vis[maxn];
void dij(int S){
memset(vis, 0, sizeof(vis));
while(!q.empty()){
P p = q.top(); q.pop();
int u = p.second;
if(vis[u]) continue;
vis[u] = 1;
for(int i = h[u] ; i != -1 ; i = e[i].next){
int w = e[i].cost, v = e[i].to;
if(dp[v][S] > dp[u][S] + w){
dp[v][S] = dp[u][S] + w;
q.push(mk(dp[v][S], v));
}
}
}
}
ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }
int main(){
memset(h, -1, sizeof(h));
n = read(), m = read(), k = read(), x = read(), y = read();
id[x] = k + 1, id[y] = k + 2; k += 2;
int tot = k;
for(int i = 1 ; i <= n ; ++i){
a[i] = read();
if(i == x || i == y) continue;
if(a[i]) id[i] = a[i];
else id[i] = ++tot;
}
n = tot;
int u, v, w;
for(int i = 1 ; i <= m ; ++i){
u = read(), v = read(), w = read();
add(id[u], id[v], w), add(id[v], id[u], w);
}
memset(dp, 0x3f, sizeof(dp));
for(int i = 1 ; i <= k ; ++i) dp[i][1 << (i - 1)] = 0;
for(int S = 1 ; S < (1 << k) ; ++S){
for(int i = 1 ; i <= n ; ++i){
for(int sub = S & (S - 1) ; sub ; sub = S & (sub - 1)){
dp[i][S] = min(dp[i][S], dp[i][sub] + dp[i][S ^ sub]);
}
if(dp[i][S] != 0x3f3f3f3f) q.push(mk(dp[i][S], i));
}
dij(S);
}
printf("%d\n", dp[1][(1 << k) - 1]);
return 0;
}
标签:ch,sub,76,int,vis,斯坦纳,练习赛,maxn,dp 来源: https://www.cnblogs.com/tuchen/p/14320421.html