学步园

这两题完全一个意思，都是差分约束的水题

找到约束条件然后建边求最短路即可

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <vector>  
#include <queue>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  

using namespace std;

const int N = 1010;
const int M = 20010;
const int inf = 0x3f3f3f3f;

queue <int>qu; 
int tot, n, m;
int head[N];
int num[N];
int dist[N];
int in_queue[N];

struct node
{
	int weight;
	int next;
	int to;
}edge[M];

void addedge(int from, int to, int weight)
{
	edge[tot].weight = weight;
	edge[tot].to = to;
	edge[tot].next = head[from];
	head[from] = tot++;
}

bool spfa(int v0)
{
	memset (dist, inf, sizeof(dist) );
	memset (in_queue, 0, sizeof(in_queue) );
	dist[v0] = 0;
	in_queue[v0]++;
	while ( !qu.empty() )
	{
		qu.pop();
	}
	qu.push(v0);
	while ( !qu.empty() )
	{
		int u = qu.front();
		qu.pop();
		for (int i = head[u]; ~i; i = edge[i].next)
		{
			int v = edge[i].to;
			if (dist[v] > dist[u] + edge[i].weight)
			{
				dist[v] = dist[u] + edge[i].weight;
				in_queue[v]++;
				if (in_queue[v] == n + 1)
				{
					return false;
				}
				qu.push(v);
			}
		}
	}
	return true;
}

int main()
{
	int  m1, m2;
	int t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d%d", &n, &m1, &m2);
		memset (head, -1, sizeof(head) );
		tot = 0;
		int u, v, w;
		while (m1--)
		{
			scanf("%d%d%d", &u, &v, &w);
			addedge(u, v, w);
		}
		while (m2--)
		{
			scanf("%d%d%d", &u, &v, &w);
			addedge(v, u, -w);
		}
		for (int i = 1; i <= n; ++i)
		{
			addedge(i + 1, i, 0);
		}
		bool flag = spfa(1);
		if (!flag)
		{
			printf("-1\n");
			continue;
		}
		for (int i = 1; i <= n; ++i)
		{
			if (dist[i] != inf)
			{
				continue;
			}
			flag = false;
			break;
		}
		if (!flag)
		{
			printf("-2\n");
			continue;
		}
		printf("%d\n", dist[n]);
	}
	return 0;
}