POJ 1192 限制条件最短路

来源：互联网发布：easing.js 编辑：程序博客网时间：2024/05/21 13:57

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;}