POJ 1192 限制条件最短路

通道在一定的时间开放，处理一下就是最短路了

strtok(char[], " ")还是很好用的

引用一下百度百科

strtok()用来将字符串分割成一个个片段。参数s指向欲分割的字符串，参数delim则为分割字符串中包含的所有字符。当strtok()在参数s的字符串中发现参数delim中包涵的分割字符时,则会将该字符改为\0 字符。在第一次调用时，strtok()必需给予参数s字符串，往后的调用则将参数s设置成NULL。每次调用成功则返回指向被分割出片段的指针。

for (i = 0, p = strtok(str, " "); p != NULL; ++i)
{
a[i] = atoi(p);
p = strtok(NULL, " ");
}

 
atof(将字符串转换成浮点型数)

atoi(将字符串转换成整型数)

atol(将字符串转换成长整型数)

strtod(将字符串转换成浮点数)

strtol(将字符串转换成长整型数)

strtoul(将字符串转换成无符号长整型数)

SFPA

#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef  long long LL;
const double PI = acos(-1.0);

template <class T> inline  T MAX(T a, T b){if (a > b) return a;return b;}
template <class T> inline  T MIN(T a, T b){if (a < b) return a;return b;}

const int N = 111;
const int M = 11111;
const LL MOD = 1000000007LL;
const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};
const int INF = 0x3f3f3f3f;

struct node
{
int w, next, v, ways;
int ti[20][2];
}edge[1111];

int cnt, head[55];
int n, m, s, t;
char str[1111];
int dist[55];
bool inq[55];

void add()
{
int a[45];
char *p;
int i, j, k, u, v;
for (i = 0, p = strtok(str, " "); p != NULL; ++i)
{
a[i] = atoi(p);
p = strtok(NULL, " ");
}
edge[cnt].w = a[2];
edge[cnt].v = a[1];
edge[cnt].next = head[a[0]];
a[2] = 0;
for (j = 3, k = 0; j < i; j +=2, ++k)
{
edge[cnt].ti[k][0] = a[j - 1];
edge[cnt].ti[k][1] = a[j];
}
if (i == j)
{
edge[cnt].ti[k][0] = a[j - 1];
edge[cnt].ti[k][1] = INF;
k++;
}
edge[cnt].ways = k;
head[a[0]] = cnt++;

edge[cnt] = edge[cnt - 1];
edge[cnt].v = a[0];
edge[cnt].next = head[a[1]];
head[a[1]] = cnt++;
}

void solve()
{
fill(dist, dist + n + 1, INF);
memset(inq, false, sizeof(inq));
dist[s] = 0;
queue < int > q;
q.push(s);
while (!q.empty())
{

int u = q.front(); q.pop(); inq[u] = false;
int i, j, k;
for (i = head[u]; i != -1; i = edge[i].next)
{
for (k = 0; k < edge[i].ways; ++k)
{
if (edge[i].ti[k][1] >= MAX(edge[i].ti[k][0], dist[u]) + edge[i].w)
{
if (MAX(edge[i].ti[k][0], dist[u]) + edge[i].w < dist[edge[i].v])
{
dist[edge[i].v] = MAX(edge[i].ti[k][0], dist[u]) + edge[i].w;
if (!inq[edge[i].v]) {inq[edge[i].v] = true; q.push(edge[i].v);}
}
}
}
}
}
if (dist[t] == INF) printf("*\n");
else printf("%d\n", dist[t]);
}

int main()
{
while (scanf("%d", &n) != EOF && n)
{
scanf("%d%d%d", &m, &s, &t);
int i, u, v;
cnt = 0;
memset(head, -1, sizeof(head));
getchar();
for (i = 0; i < m; ++i)
{
gets(str);
add();
}
if (s == t) printf("0");
else solve();
}
return 0;
}