【题解】LuoGu1841：[JSOI2007]重要的城市

原题传送门
$floyd$算出两两之间最短路$dist$以及最短路方案数$num$
对于两个城市$j,k$，$i$是$j到k$最短路非走不可的城市的充要条件是$dis{t}_{j,k}=dis{t}_{j,i}+dis{t}_{i,k}且nu{m}_{j,k}=nu{m}_{j,i}\ast nu{m}_{i,k}$
时间$O(n^3)$

Code：

#include 
#define maxn 210
#define LL long long
using namespace std;
int w[maxn][maxn], n, m;
LL num[maxn][maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
} 

int main(){
	n = read(), m = read();
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j) w[i][j] = 1e9;
	for (int i = 1; i <= m; ++i){
		int x = read(), y = read(), z = read();
		w[x][y] = w[y][x] = min(w[x][y], z);
		num[x][y] = num[y][x] = 1;
	}
	for (int k = 1; k <= n; ++k)
		for (int i = 1; i <= n; ++i)
			if (i != k)
				for (int j = 1; j <= n; ++j)
					if (i != j && j != k){
						if (w[i][j] > w[i][k] + w[k][j]) w[i][j] = w[i][k] + w[k][j], num[i][j] = num[i][k] * num[k][j]; else
						if (w[i][j] == w[i][k] + w[k][j]) num[i][j] += num[i][k] * num[k][j];
					}
//	printf("%d\n", num[1][3]);
	int Flag = 0;
	for (int i = 1; i <= n; ++i){
		int flag = 0;
		for (int j = 1; j <= n; ++j){
			if (flag) break;
			if (i != j)
				for (int k = 1; k <= n; ++k)
					if (i != k && j != k)
					if (w[j][k] == w[j][i] + w[i][k] && num[j][k] == num[j][i] * num[i][k]){
						printf("%d ", i);
						flag = 1, Flag = 1;
						break;
					}
		}
	}
	if (!Flag) puts("No important cities.");
	return 0;
}