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jzoj 1328

题目大意

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 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3

输出样例

10

样例解释

数据范围

$1⩽x⩽N⩽1000$
$1⩽m⩽100000$
$1⩽t_i⩽100$

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;
}