hdu 4085 Peach Blossom Spring

Peach Blossom Spring

题意：有n个城市，m条路，[1,k]的城市有居民， [n-k+1, n]的城市有庇护所， 现在要修路， 使得每一座城市的居民都可以到达一个庇护所， 并且一个庇护所只能容纳一个城市的居民， 现在求所有城市的居民都能到达庇护所的最小花费。

题解：斯坦纳树跑出花费。 d[i][state] = cost， i代表的是第i个城市， state代表的是联通的状态， val代表的是花费。

先跑出所有的d[i][state]的花费。然后由于不需要是联通图， 再最后跑出每种符合的最小花费。

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+7;
const int N = 55, M = 2005;
int head[N], nt[M], to[M], ct[M], tot;
int d[N][1<<10], s[N], vis[N][1<<10], dp[1<<10];
int n, m, k, nn;
void init(){
    memset(head, -1, sizeof(head));
    memset(s, 0, sizeof(s));
    tot = 0;
}
inline void add(int u, int v, int w){
    to[tot] = v;
    ct[tot] = w;
    nt[tot] = head[u];
    head[u] = tot++;
}
void input(){
    int u, v, c;
    for(int i = 1; i <= m; i++){
        scanf("%d%d%d", &u, &v, &c);
        add(u, v, c);
        add(v, u, c);
    }
    for(int i = 1; i <= n; i++)
        for(int j = 0; j < nn; j++) d[i][j] = inf;
    for(int i = 1; i <= k; i++){
        s[i] = 1<<(i-1); d[i][s[i]] = 0;
        s[n-i+1] = 1<<(k+i-1), d[n-i+1][s[n-i+1]] = 0;
    }
}
queue<int> q;
inline bool update(int x, int y, int w){
    if(d[x][y] <= w) return false;
    d[x][y] = w;
    return true;
}
void spfa(){
    int t, x, y;
    while(!q.empty()){
        t = q.front();
        q.pop();
        x = t / 10000, y = t%10000;
        vis[x][y] = 0;
        for(int i = head[x]; ~i; i = nt[i]){
            if(update(to[i],y|s[to[i]],d[x][y]+ct[i]) && y == (y|s[to[i]]) && !vis[to[i]][y]){
                vis[to[i]][y] = 1;
                q.push(to[i]*10000+y);
            }
        }
    }
}
bool check(int x){
    int r = 0;
    for(int i = 0;x;i++, x>>=1)
        r += (x&1)*(i<k?1:-1);
    return r==0;
}
void solve(){
    for(int i = 0; i < nn; i++){
        for(int j = 1; j <= n; j++){
            for(int k = (i-1)&i; k; k = (k-1)&i)
                d[j][i] = min(d[j][i], d[j][k|s[j]]+d[j][s[j]|(i-k)]);
            if(d[j][i] < inf) q.push(i+j*10000), vis[j][i] = 1;
        }
        spfa();
    }
    for(int i = 0; i < nn; i++){
        dp[i] = inf;
        for(int j = 1; j <= n; j++) dp[i] = min(dp[i], d[j][i]);
    }
    for(int i = 1; i < nn; i++){
        if(check(i)){
            for(int j = i&(i-1); j; j = (j-1)&i){
                if(check(j)) dp[i] = min(dp[i], dp[j]+dp[i-j]);
            }
        }
    }
    if(dp[nn-1]>=inf) puts("No solution");
    else printf("%d
", dp[nn-1]);
}
int main(){
    int t, u, v, c;
    scanf("%d", &t);
    while(t--){
        init();
        scanf("%d%d%d", &n, &m, &k);
        nn = 1<<(2*k);
        input();
        solve();
    }
    return 0;
}