HDU ~ 3592 ~ World Exhibition （SPFA + 差分约束）

和上一道题一样：Layout

只不过是T组输入输出，样例变了一下。

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 1e6 + 5;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to, dist; //起点,终点,距离
Edge(int u, int v, int w):from(u), to(v), dist(w) {}
};
struct BellmanFord
{
int n, m; //结点数,边数(包括反向弧)
vector<Edge> edges; //边表。edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
bool vis[MAXN]; //是否在队列中
int d[MAXN]; //Bellman-Ford
int p[MAXN]; //上一条弧
int cnt[MAXN]; //进队次数

void init(int n)
{
this->n = n;
edges.clear();
for (int i = 0; i <= n; i++) G[i].clear();
}
void add_edge(int from, int to, int dist)
{
edges.push_back(Edge(from, to, dist));
//edges.push_back((Edge){from, to, dist}); G++编译通过，C++可能会编译错误
m = edges.size();
G[from].push_back(m - 1);
}

bool bellman_ford(int s)
{
for (int i = 0; i <= n; i++) d[i] = INF;
memset(vis, 0, sizeof(vis));
memset(cnt, 0, sizeof(cnt));
d[s] = 0; vis[s] = true;

queue<int> Q;
Q.push(s);
while (!Q.empty())
{
int u = Q.front(); Q.pop();
vis[u] = false;
for (int i = 0; i < G[u].size(); i++)
{
Edge& e = edges[G[u][i]];
if (d[u] < INF && d[e.to] > d[u] + e.dist)
{
d[e.to] = d[u] + e.dist;
p[e.to] = G[u][i];
if (!vis[e.to])
{
Q.push(e.to); vis[e.to] = true;
if (++cnt[e.to] > n) return false;//有负环
}
}
}
}
return true;//没有负环
}
};
BellmanFord solve;
int main()
{
int T, N, ML, MD;
scanf("%d", &T);
while(T--)
{
scanf("%d%d%d", &N, &ML, &MD);
solve.init(N);
while(ML--)
{
int u, v, t;
scanf("%d%d%d", &u, &v, &t);
solve.add_edge(u, v, t);
}
while (MD--)
{
int u, v, t;
scanf("%d%d%d", &u, &v, &t);
solve.add_edge(v, u, -t);
}
for (int i = 2; i <= N; i++) solve.add_edge(i, i - 1, -1);

if (solve.bellman_ford(1))
{
if (solve.d
== INF) printf("-2\n");
else printf("%d\n", solve.d
);
}
else printf("-1\n");
}
return 0;
}
/*
1
4 2 1
1 3 8
2 4 15
2 3 4
*/