有一个有向图，给你一个x，让你求每一个点到x再回去的最短路径，输出所有最短路径的最大值
输入样例 4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 2 3 输出样例10
样例解释 数据范围1⩽x⩽N⩽10001\leqslant x\leqslant N\leqslant 10001⩽x⩽N⩽1000 1⩽m⩽1000001\leqslant m\leqslant 1000001⩽m⩽100000 1⩽ti⩽1001\leqslant t_i\leqslant 1001⩽ti⩽100
解题思路每一条边建一条正边一条反边，从x跑两遍spfa，然后合在一起即可
代码 #include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int n, m, s, x, y, z, w, h, ans, tot, p[1500], b[1500], bb[1500], head[1500], headd[1500];struct rec{int to, l, next;}a[100500], e[100500];void add(int x, int y, int z){a[++tot].to = y;a[tot].l = z;a[tot].next = head[x];head[x] = tot;e[++w].to = x;//反边e[w].l = z;e[w].next = headd[y];headd[y] = w;}void spfa(){memset(b, 127/3, sizeof(b));queue<int>d;d.push(s);p[s] = 1;b[s] = 0;while(!d.empty())//最短路{h = d.front();d.pop();for (int i = head[h]; i; i = a[i].next)if (b[h] + a[i].l < b[a[i].to]){b[a[i].to] = b[h] + a[i].l;if (!p[a[i].to]){p[a[i].to] = 1;d.push(a[i].to); }} p[h] = 0;} memset(bb, 127/3, sizeof(bb));d.push(s);p[s] = 1;bb[s] = 0;while(!d.empty())//反的最短路{h = d.front();d.pop();for (int i = headd[h]; i; i = e[i].next)if (bb[h] + e[i].l < bb[e[i].to]){bb[e[i].to] = bb[h] + e[i].l;if (!p[e[i].to]){p[e[i].to] = 1;d.push(e[i].to); }} p[h] = 0;} return;}int main(){scanf("%d %d %d", &n, &m, &s);for (int i = 1; i <= m; ++i){scanf("%d %d %d", &x, &y, &z);add(x, y, z);} spfa();for (int i = 1; i <= n; ++i)ans = max(ans, b[i] + bb[i]);//合在一起printf("%d", ans);return 0;}
