ACM/ICPC 之 Bellman Ford练习题(ZOJ1791(POJ1613))

这道题稍复杂一些，需要掌握字符串输入的处理+限制了可以行走的时间。

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ZOJ1791(POJ1613)-Cave Raider

//限制行走时间的最短路
//POJ1613-ZOJ1791
//Time:16Ms	Memory:324K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

#define MAX	505
#define MAXT 55
#define MAXS MAXT*3
#define INF 0x3f3f3f3f

struct Edge {
	int u, v, w;
	int t[MAXT], lt;	//lt:开启与关闭时间点总和
}e[MAX];

int n, m, s, t;
int d[MAXT];

void bellman_ford()
{
	memset(d, 0x3f, sizeof(d));
	d[s] = 0;
	for (int i = 1; i <= n; i++)
		for (int j = 0; j < m; j++)
			for (int k = 1; k <= e[j].lt; k += 2)
			{
				int u = e[j].u, v = e[j].v;
				//tu:从u出发到v的时间
				int tu = max(d[u], e[j].t[k - 1]) + e[j].w;
				int tv = max(d[v], e[j].t[k - 1]) + e[j].w;
				if (tu <= e[j].t[k] || tv <= e[j].t[k])	//几次WA是因为没有取'='
				{
					if (tv <= e[j].t[k]) d[u] = min(d[u], tv);
					if (tu <= e[j].t[k]) d[v] = min(d[v], tu);
					break;
				}
			}
}

int main()
{
	char str[MAXS];
	while (scanf("%d", &n), n)
	{
		memset(e, 0, sizeof(e));
		scanf("%d%d%d", &m, &s, &t);
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
			e[i].lt = 1;
			cin.getline(str, MAXS);
			int len = strlen(str);
			for (int j = 0; j < len; j++)
			{
				bool flag = false;	//有无数值
				while (str[j] >= '0' && str[j] <= '9')
				{
					e[i].t[e[i].lt] = e[i].t[e[i].lt] * 10 + str[j++] - '0';
					flag = true;	//已记录数值
				}
				if (flag)	e[i].lt++;
			}
			e[i].t[e[i].lt] = INF;	//偶数时当做+∞，奇数时无用
		}

		bellman_ford();
		if (d[t] == INF) printf("*\n");
		else	printf("%d\n", d[t]);
	}
	return 0;
}