PAT 甲级 1087 All Roads Lead to Rome（SPFA+DP）

题目链接 All Roads Lead to Rome

题目大意：求符合题意（三关键字）的最短路。并且算出路程最短的路径有几条、

思路：求最短路并不难，SPFA即可，关键是求总路程最短的路径条数。

我们令$h[i][j]$为$i$到$j$的最短路，$g[i][j]$为$i$到$j$的最短路条数。

$f[i][j]$的求法就是常规的Floyd求法

$g[i][j]$利用乘法原理的性质来求。

当枚举的中间点$k$满足 $f[i][k] + f[k][j] = f[i][j]$时

　　$g[i][j] += g[i][k] * g[k][j]$

当$f[i][k] + f[k][j] < f[i][j]$时

　　$g[i][j] = g[i][k] * g[k][j]$

时间复杂度 $O(kn + n^{3})$

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)

const int N = 210;

int g[N][N], h[N][N], c[N], d[N], e[N], f[N], pp[N];
int n, m, s, t, x;
bool inqueue[N];
string s1, s2, b[N];
map <string, int> mp;
vector < pair <int, int> > v[N];
queue <int> Q;

void print(int x){
	if (pp[x] == 0){  cout << b[x]; return ; }
	else print(pp[x]);
	cout << "->" << b[x];
}


int main(){

	scanf("%d%d", &n, &m);
	cin >> s1;

	s = 1; mp[s1] = 1; c[1] = 0; b[1] = s1;
	rep(i, 2, n){
		cin >> s2 >> x;
		mp[s2] = i;
		b[i] = s2;
		c[i] = x;
		if (s2 == "ROM") t = i;
	}

	
	rep(i, 1, n) rep(j, 1, n) h[i][j] = 1 << 28, g[i][j] = 0;
	rep(i, 1, n) h[i][i] = 0;

	rep(i, 1, m){
		cin >> s1 >> s2 >> x;
		int a1 = mp[s1], a2 = mp[s2];

		v[a1].push_back({a2, x});
		v[a2].push_back({a1, x});

		if (h[a1][a2] > x){
			h[a1][a2] = h[a2][a1] = x;
			g[a1][a2] = g[a2][a1] = 1;
		}

		else if (h[a1][a2] == x){
			++g[a1][a2];
			++g[a2][a1];
		}
	}

	rep(k, 1, n){
		rep(i, 1, n){
			rep(j, 1, n){
				if (h[i][k] + h[k][j] == h[i][j]) g[i][j] += g[i][k] * g[k][j];
				else if (h[i][k] + h[k][j] < h[i][j]){
					h[i][j] = h[i][k] + h[k][j];
					g[i][j] = g[i][k] * g[k][j];
				}
			}
		}
	}
		

	memset(inqueue, false, sizeof inqueue);

	Q.push(s);
	rep(i, 1, n) d[i] = 1 << 28;
	rep(i, 1, n) e[i] = 1 << 28;
	rep(i, 1, n) f[i] = 0;
	d[s] = 0; e[s] = 0; f[s] = 0;
	rep(i, 1, n) pp[i] = 0;

	while (!Q.empty()){
		int x = Q.front(); Q.pop();
		inqueue[x] = false;
		for (auto u : v[x]){
			int p = u.first, w = u.second;
			if (d[x] + w < d[p] || d[x] + w == d[p] && f[x] + c[p] > f[p] 
					|| d[x] + w == d[p] && f[x] + c[p] == f[p] && e[x] + 1 < e[p]){
				d[p] = d[x] + w;
				f[p] = f[x] + c[p];
				e[p] = e[x] + 1;
				pp[p] = x;
				if (!inqueue[p]){
					inqueue[p] = true;
					Q.push(p);
				}
			}
		}
	}

	printf("%d %d %d %d
", g[s][t], d[t], f[t], f[t] / e[t]);
	print(t);
	putchar(10);
	return 0;

}