同余最短路

题意大致就是 一张无向图  每条边可以多次经过  问s,到t 的不小于k的最短路  （或者仅仅等于k的最短路）

取m=2*min（s到下一个点的最短路径）:   如果一条路径长度sum可行   那么sum+m也是可行的   

然后 dis[i][j]表示从起点s到i点 ，路径长度%m==j  的最短路径    

最后就是利用同余的性质去跑最短路spfa

题目  HDU 6071

          51NOD 1326

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>  
#include <iostream>  
#include <cstdlib>  
#include <cstring>  
#include <string>
#include <cstdio>  
#include <climits>
#include <cmath> 
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <sstream>
#define LL long long
#define fora(i,a,n) for(int i=a;i<=n;i++)
#define fors(i,n,a) for(int i=n;i>=a;i--)
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%lld",&x)
const int MAXN = 100024;
const long long INF = 1e18 + 60002;
const double eps = 1e-8;
using namespace std;
struct Node {
	LL v, mod, diss;
	Node(LL _v, LL _mod, LL _diss) {
		v = _v; mod = _mod; diss = _diss;
	}
	Node() {}
};
LL t, k, m;
LL g[4][60003];
bool vis[4][60003];
LL dis[4][4];
void spfa(LL v) {
	queue<Node>q;
	memset(vis, 0, sizeof(vis));
	memset(g, 0x3f, sizeof(g));

	q.push(Node(1, 0, 0));
	vis[1][0] = 1;

	LL nex_dis, nex_mod, nex_v;
	while (!q.empty()) {
		Node f = q.front(); q.pop();
		vis[f.v][f.mod] = 0;
		for (int i = -1, j = 0; j <= 1; i += 2, j++) {
			nex_v = (f.v + i + 4) % 4;
			nex_dis = f.diss + dis[f.v][nex_v];
			nex_mod = nex_dis%m;
			if (g[nex_v][nex_mod]>nex_dis) {
				g[nex_v][nex_mod] = nex_dis;
				if(!vis[nex_v][nex_mod]){
					vis[nex_v][nex_mod] = 1;
					q.push(Node(nex_v, nex_mod, nex_dis));
				}
			}
		}
	}
	LL Min = 4*INF, num;
	for (int i = 0; i<m; i++) {
		num = g[1][i];
		if (num<k) { num += ((k - num - 1) / m + 1)*m; }
		if (num<Min)
			Min = num;
	}
	printf("%lld\n", Min);
}
int main() {
#ifdef local
	freopen("ini.txt", "r", stdin);
	//freopen("out.txt","w",stdout);
#endif
	scanf("%lld", &t);
	while (t--) {
		memset(dis, 0, sizeof(dis));
		scanf("%lld", &k);
		for (int i = 0; i<4; i++) scanf("%lld", &dis[i][(i + 1) % 4]), dis[(i + 1) % 4][i] = dis[i][(i + 1) % 4];
		m = 2 * min(dis[1][0], dis[1][2]);
		spfa(1);
	}
	return 0;
}