SGU 226.Colored graph

时间限制：0.25s

空间限制：4M

题意：

给出一个n个节点，m条边的图，每条边都有标记了编号为1,2,3三种颜色之一，现在求从1号节点到n号节点的一条最短路径的长度，要求该路径中相邻的边没有相同的颜色。

有限制条件的SPFA，要注意有时要形成环来改变路径颜色，才能到达目标点。

参考代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define INF 300
using namespace std;
struct node {
	int v, ne, c;
} edge[INF * INF];
queue<int> ql;
int head[INF], pd[INF];
int dis[INF][4], cnt,n,m;
void added (int u, int v, int c) {
	edge[++cnt].v = v, edge[cnt].c = c;
	edge[cnt].ne = head[u];
	head[u] = cnt;
}
void spfa() {
	while (!ql.empty() ) {
		int x = ql.front();
		pd[x] = 0,ql.pop();
		for (int i = head[x]; i != 0; i = edge[i].ne) {
			int j = edge[i].v, color = edge[i].c;
			for (int k = 1; k <= 3; k++) {
				if (k == color || dis[x][k] == -1) continue;
				if (dis[j][color] == -1 || dis[j][color] > dis[x][k] + 1) {
					dis[j][color] = dis[x][k] + 1;
					if (!pd[j])
						pd[j] = 1, ql.push (j);
				}
			}
		}
	}
}
int main() {
	int  x, y, c;
	scanf ("%d %d", &n, &m);
	memset (dis, -1, sizeof dis);
	for (int i = 1; i <= m; i++) {
		scanf ("%d%d%d", &x, &y, &c);
		added (x, y, c);
	}
	ql.push (1); pd[1] = 1;
	dis[1][1] = dis[1][2] = dis[1][3] = 0;
	spfa();
	int ans = INF<<12;
	for (int i = 1; i <= 3; i++)
		if (dis[n][i] != -1)
			ans = min (ans, dis[n][i]);
	if (ans != INF<<12) printf ("%d", ans);
	else
		puts ("-1");
	return 0;
}

SGU 226.Colored graph