Solution
参考官方大大的题解
#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int N = 1e3 + 7; //节点数
const int M = 3e3 + 7; //路径数
int head[N], tot = 0;//前向星变量
struct Node {
int u; //起点
int w; //权值
int v, next;
} edge[M << 1];
void add(int u, int v, int w) {
tot++;
edge[tot].u = u;
edge[tot].v = v;
edge[tot].w = w;
edge[tot].next = head[u];
head[u] = tot;
}
bool vis[N];
ll dis[N];
void spfa() {
ms(vis, 0); ms(dis, 0x3f);
queue<int> q;
q.push(0); dis[0] = 0;
while (q.size()) {
int u = q.front(); q.pop();
vis[u] = 0;
for (int i = head[u]; i; i = edge[i].next) {
int v = edge[i].v, w = edge[i].w;
if (dis[v] > dis[u] + w) {
dis[v] = dis[u] + w;
if (!vis[v]) { q.push(v), vis[v] = 1; }
}
}
}
}
int main() {
int n = read(), m = read();
for (int i = 1; i <= m; ++i) {
int u = read(), v = read(), w = read();
add(u, v, w);
}
for (int i = 1; i <= n; ++i) add(0, i, 0);
spfa();
for (int i = 1; i <= m; ++i) {
int u = edge[i].u, v = edge[i].v, w = edge[i].w;
printf("%d %d %lld\n", u, v, dis[u] - dis[v] + w);
}
return 0;
}