这道题难得不是算法，而是处理。
题意就是让你求最短路，只有当两个点在某一秒颜色相同时，这条边才可以通行，输入首先给你 起点和终点， 然后给你 点数和边数，
接下来 n 行 初始颜色，初始颜色持续时间，蓝色持续时间，紫色持续时间。 再接下来m行，无向边的起点和终点以及通过所需的时间。
题意他说的有些模糊，样例我看了很多遍也不对，后来才发现如果你在某一秒到达一个点，这一秒颜色和下一个点相同，但是下一秒这个点就要变色，那么是不能在这一秒走的，这个具体处理起来很麻烦
这篇博客说的很详细，戳链接：http://www.cnblogs.com/Rinyo/archive/2012/11/29/2795030.html
上代码……
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#define N 350
#define M 15000*2
#define inf 1<<30
using namespace std;int n, m, S, T;
int p[N] = {0}, next[M], v[M], bnum = 0, c[M], fa[N];
int dis[N], vis[N], firstcolor[N], firstremain[N], remain[N][2];void addbian(int x, int y, int z)
{
bnum++; next[bnum] = p[x]; p[x] = bnum;
v[bnum] = y; c[bnum] = z;
bnum++; next[bnum] = p[y]; p[y] = bnum;
v[bnum] = x; c[bnum] = z;
}void calt(int now, int nowtime, int &color, int &changetime)
{
if (nowtime < firstremain[now])
{
color = firstcolor[now];
changetime = firstremain[now];
return;
}
int k;
k = (nowtime-firstremain[now]) % (remain[now][0]+remain[now][1]);
nowtime -= k;
if (firstcolor[now])
{
if (k < remain[now][0]) { color = 0; changetime = nowtime + remain[now][0]; }
else { color = 1; changetime = nowtime + remain[now][1] + remain[now][0]; }
}
else
{
if (k < remain[now][1]) { color = 1; changetime = nowtime + remain[now][1]; }
else { color = 0; changetime = nowtime + remain[now][0] + remain[now][1]; }
}
}int gettime(int x, int y, int nowtime, int dis, int f)
{
int ca, cb, ta, tb, ta1, ca1;
calt(x, nowtime, ca, ta);
calt(y, nowtime, cb, tb);
if (ca == cb) return nowtime + dis;
if (ta == tb)
{
if (f == 0) return gettime(x, y, ta, dis, f+1);
else if (nowtime <= firstremain[x] || nowtime <= firstremain[y])
gettime(x, y, ta, dis, f+1);
else return inf;
}
return min(ta, tb) + dis;
}void spfa()
{
queue<int> q;
for (int i = 1; i <= n; ++i) { vis[i] = 0; dis[i] = inf; }
q.push(S); dis[S] = 0; vis[S] = 1;
while (!q.empty())
{
int now = q.front(); q.pop();
int k = p[now];
while (k != 0)
{
int t = gettime(now, v[k], dis[now], c[k], 0);
if (dis[v[k]] > t)
{
dis[v[k]] = t; fa[v[k]] = now;
if (!vis[v[k]])
{
vis[v[k]] = 1;
q.push(v[k]);
}
}
k = next[k];
}
vis[now] = 0;
}
}void print(int now)
{
if (now == S) printf("%d ", S);
else
{
print(fa[now]);
if (now != T) printf("%d ", now);
else printf("%d\n", T);
}
}int main()
{
scanf("%d%d", &S, &T);
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i)
{
char s[2];
scanf("%s", s);
if (s[0] == ‘B‘) firstcolor[i] = 0;
else firstcolor[i] = 1;
scanf("%d%d%d", &firstremain[i], &remain[i][0], &remain[i][1]);
}
for (int i = 1; i <= m; ++i)
{
int x, y, z; scanf("%d%d%d", &x, &y, &z);
addbian(x, y, z);
}
spfa();
if (dis[T] == inf)
{
printf("0\n");
return 0;
}
printf("%d\n", dis[T]);
print(T);
}
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