I

- 题目大意

    可以给每个点的入边加一个值和出边加一个值，问最小的边权最大是多少。

- 解题思路

    根据题目的描述，可以列出一个不等式d(a,b) +x(a)-x(b)>=m，移项可得x(b)-x(a)<=d(a,b)-m正好满足差分约束的形式。所有的边就对应着一个差分约束系统。差分约束有解的充要条件是不存在负环。然后利用SPFA和二分法来查找即可。

- 代码

#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstring>
using namespace std;
const int N = 1e3;
const int M = 1e6 + 5;
const int INF = 0x3f3f3f;
int n, m;
struct edge {
	int v, next, w;
}e[M * 2];
int head[N], cnt;
int d[N];
int inq[N];
int cn[N];
void addedge(int u, int v, int c)
{
	e[cnt].w = c;
	e[cnt].v = v;
	e[cnt].next = head[u];
	head[u] = cnt++;
}
bool SPFA(int s) {
	memset(d, INF, sizeof(d));
	memset(inq, 0, sizeof(inq));
	memset(cn, 0, sizeof(cn));
	queue<int> Q;
	Q.push(s);
	d[s] = 0;
	inq[s] = 1;
	cn[s]++;
	while (Q.size()) {
		int u = Q.front();
		Q.pop();
		inq[u] = 0;
		for (int i = head[u]; ~i; i = e[i].next)
		{
			int v = e[i].v;
			int w = e[i].w;
			if (d[v] > d[u] + w)
			{
				d[v] = d[u] + w;
				if (!inq[v])
				{
					Q.push(v);
					inq[v] = 1;
					if (++cn[v] >= n+1)
						return false;
				}
			}
		}
	}
	return true;
}

bool check(int x)
{
	bool vis = true;
	for (int i = 0; i <= n; i++)
	{
		for (int j = head[i]; j != -1; j = e[j].next)
			e[j].w -= x;
	}
	if (!SPFA(0))
		vis = false;
	for (int i = 0; i <= n; i++)
	{
		for (int j = head[i]; j != -1; j = e[j].next)
			e[j].w += x;
	}
	return vis;
}

int main()
{
	int a, b, c;
	while(~scanf("%d%d", &n, &m))
    {
    int sum = 1;
	cnt = 0;
	memset(head, -1, sizeof(head));
	int l = 1, r = -INF, mid;

	for (int i = 0; i < m; i++)
	{
		scanf("%d%d%d", &a, &b, &c);
		addedge(a, b, c);
		r = max(r, c);
	}
	for (int i = 1; i <= n; i++)
		addedge(0, i, 0);
	if (check(r + 1))
	{
		printf("Infinite
");
		continue;
	}
	else if (!check(1))
	{
		printf("No Solution
");
		continue;
	}
	else
	{
		while (r - l >= 0)
		{
			mid = (r + l) / 2;
			if (check(mid))
			{
				l = mid + 1;
				sum = mid;
			}
			else
			{
				r = mid - 1;
			}
		}
	}
	printf("%d
", sum);
    }
	return 0;
}