题解【洛谷P1938】 [USACO09NOV]找工就业Job Hunt
题解
将路径连边\((x, y, d)\) ,将航线连边\((x, y, d - w)\)。其中线路是从\(x\)到\(y\),航线的费用为\(w\),\(d\)的含义如题面。
跑一遍\(SPFA\)最长路即可。
注意判断负环的情况,此时要输出\(-1\)。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <queue>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return f * x;
}
int d, p, c, f, s, tot, head[1003], nxt[1003], ver[1003], edge[1003], in[1003], ans, dis[1003];
int vis[1003];
queue <int> q;
inline void add(int u, int v, int w)
{
ver[++tot] = v, edge[tot] = w, nxt[tot] = head[u], head[u] = tot;
}
inline void SPFA()
{
q.push(s);
vis[s] = 0;
dis[s] = d;
in[s] = 1;
while (!q.empty())
{
int u = q.front(); q.pop();
vis[u] = 0;
++in[u];//统计入队次数
if (in[u] > c) {puts("-1"); exit(0);}//负环
for (int i = head[u]; i; i = nxt[i])
{
int v = ver[i], w = edge[i];
if (dis[v] < dis[u] + w)//注意是小于号
{
dis[v] = dis[u] + w;
if (!vis[v])
{
vis[v] = 1;
q.push(v);
}
}
}
}
}
int main()
{
//File("P1938");
d = gi(), p = gI(), c = gI(), f = gI(), s = gI();
for (int i = 1; i <= p; i+=1)
{
int u = gI(), v = gI();
add(u, v, d);//连边
}
for (int i = 1; i <= f; i+=1)
{
int u = gi(), v = gI(), w = gi();
add(u, v, d - w);//连边
}
SPFA();//SPFA最长路
int maxx = 0;
for (int i = 1; i <= c; i+=1) maxx = max(maxx, dis[i]);
printf("%d\n", maxx);
return 0;
}