题解 变异大老鼠

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
#define db double
#define fir first
#define sec second
#define make make_pair
//#define int long long

int n, m, k;
int back[N], cnt[N];
ll dis[N];
db p[310][310];
bool vis[N];
priority_queue<pair<ll, int>> q;
namespace edge1{
	int head[N], size;
	struct edge{int to, next, val;}e[N<<1];
	inline void add(int s, int t, int w) {e[++size].to=t; e[size].val=w; e[size].next=head[s]; head[s]=size;}
}
namespace edge2{
	int head[N], size;
	struct edge{int to, next;}e[N<<1];
	inline void add(int s, int t) {e[++size].to=t; e[size].next=head[s]; head[s]=size;}
}

void init() {
	using namespace edge1;
	memset(dis, 0x3f, sizeof(dis));
	dis[1]=0;
	q.push(make(0, 1));
	pair<ll, int> u;
	while (q.size()) {
		u=q.top(); q.pop();
		if (vis[u.sec]) continue;
		vis[u.sec]=1;
		for (int i=head[u.sec],v; ~i; i=e[i].next) {
			v = e[i].to;
			if (dis[u.sec]+e[i].val < dis[v]) {
				dis[v]=dis[u.sec]+e[i].val; back[v]=u.sec;
				q.push(make(-dis[v], v));
			}
		}
	}
	for (int i=2; i<=n; ++i) edge2::add(back[i], i), ++cnt[back[i]];
}

namespace force{
	int num[N];
	db ans;
	db dfs2(int u) {
		using namespace edge2;
		db sum=p[u][num[u]];
		db ps=1.0/cnt[u];
		for (int i=head[u],v; ~i; i=e[i].next) {
			v = e[i].to;
			sum+=ps*dfs2(v);
		}
		return sum;
	}
	void dfs1(int u, int rest) {
		if (u>n) {
			if (rest==0) ans=max(ans, dfs2(1));
			return ;
		}
		for (int i=0; i<=rest; ++i) {
			num[u]=i;
			dfs1(u+1, rest-i);
		}
	}
	void solve() {
		dfs1(1, k);
		printf("%.6lf\n", min(ans, 1.0));
		exit(0);
	}
}

namespace task1{
	db dp[310][30010], tem[310][30010];
	void dfs(int u) {
		using namespace edge2;
		// cout<<"dfs: "<<u<<' '<<cnt[u]<<endl;
		for (int i=1; i<=k; ++i) dp[u][i]=p[u][i];
		db ps=cnt[u]?(1.0/cnt[u]):1.0;
		// cout<<"ps: "<<u<<' '<<ps<<endl;
		for (int i=head[u],v; ~i; i=e[i].next) {
			v = e[i].to;
			dfs(v);
			for (int s=k; ~s; --s) {
				for (int t=k-s; ~t; --t) {
					tem[u][s+t]=max(tem[u][s+t], tem[u][s]+ps*dp[v][t]);
				}
			}
		}
		for (int s=k; ~s; --s) {
			for (int t=k-s; ~t; --t) {
				dp[u][s+t]=max(dp[u][s+t], dp[u][s]+(1-dp[u][s])*tem[u][t]);
			}
		}
	}
	void solve() {
		dfs(1);
		printf("%.6lf\n", min(dp[1][k], 1.0));
		// cout<<"---dp---"<<endl;
		// for (int i=1; i<=n; ++i) {for (int j=1; j<=k; ++j) printf("%.3lf ", dp[i][j]); cout<<endl;}
		exit(0);
	}
}

signed main()
{
	freopen("arrest.in", "r", stdin);
	freopen("arrest.out", "w", stdout);

	using namespace edge1;
	memset(edge1::head, -1, sizeof(edge1::head));
	memset(edge2::head, -1, sizeof(edge2::head));
	scanf("%d%d%d", &n, &m, &k);
	for (int i=1,x,y,z; i<=m; ++i) {
		scanf("%d%d%d", &x, &y, &z);
		add(x, y, z); add(y, x, z);
	}
	for (int i=1; i<=n; ++i) for (int j=1; j<=k; ++j) scanf("%lf", &p[i][j]);
	init();
	task1::solve();

	return 0;
}