1087 All Roads Lead to Rome (30 分)

1087 All Roads Lead to Rome (30 分)

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#include <iostream>
#include <cstring>
#include <map>
using namespace std;
const int inf = 0x3f3f3f3f;
int e[250][250], vis[250], dis[250], path[250], ret[250], cnt[250], a[250], aa[250];
double avg[250];
map<string, int> m;
map<int, string> m1;
int n, M, d, x;
void dijkstra(int s){
	memset(dis, inf, sizeof dis);
	dis[s] = 0;
	ret[s] = 1;
	for(int i = 0; i < n; ++i){
		int t = -1;
		for(int j = 1; j <= n; ++j)
			if(!vis[j] && (t == -1 || dis[j] < dis[t]))
				t = j;
		vis[t] = 1;
		for(int j = 1; j <= n; ++j) {
			if(!vis[j] && dis[j] > dis[t] + e[t][j]) {
				dis[j] = dis[t] + e[t][j];
				aa[j] = aa[t] + a[j];
				ret[j] = ret[t];
				cnt[j] = cnt[t] + 1;
				avg[j] = aa[j] * 1.0 / cnt[j];
				path[j] = t;
			}
			else if(!vis[j] && dis[j] == dis[t] + e[t][j]) {
				ret[j] += ret[t]; 
				if(aa[j] < aa[t] + a[j]){
					aa[j] = aa[t] + a[j];
					cnt[j] = cnt[t] + 1;
					avg[j] = aa[j] * 1.0 / cnt[j];
					path[j] = t;
				}
				else if(aa[j] == aa[t] + a[j] && aa[j] * 1.0 / (cnt[t] + 1) > avg[j]){
					cnt[j] = cnt[t] + 1;
					avg[j] = aa[j] * 1.0 / cnt[j];
					path[j] = t;
				}
			}
		}
	}
}
void dfs(int x) {
	if(x == 1){
		cout << m1[x];
		return;
	}
	dfs(path[x]);
	cout << "->" << m1[x];
}
int main() {
	string s1, s2;
	cin >> n >> M >> s1;
	memset(e, inf, sizeof e);
	for(int i = 1; i <= n; ++i)
		e[i][i] = 0;
	m[s1] = 1;
	m1[1] = s1;
	for(int i = 2; i <= n; ++i){
		cin >> s1 >> a[i];
		m[s1] = i;
		m1[i] = s1;
		if(s1 == "ROM") d = i;
	}
	while(M--) {
		cin >> s1 >> s2 >> x;
		e[m[s1]][m[s2]] = e[m[s2]][m[s1]] = x;
	}
	dijkstra(1);
	printf("%d %d %d %d\n", ret[d], dis[d], aa[d], int(avg[d]));
	dfs(d);
}