题意:
给出一个无向图,问从1到n是否存在一条长度为L的路径。
n,m<=50,1<=路径长度<=10000,L<=10^18
思路:
改变一下思路,我们发现,假设从起点1走到终点N有一条路径的长度为a,假设它再往一条与终点相连的长为b的路径反复走无数次后使得路径长度到达了T,那么一定有(T-a)%(2*k)==0,即**T%(2*k)=a%(2*k),所以我们只需要看是否从1到N存在一条路径长度为d,使得d%(2*k)=t%(2*k)
因为这个题目和模数有关,所以我们要把取摸的结果写入状态.
dp[i][j]表示处于i节点,从一号点到i号点的花费和%(2*k)(选择的边权)等与k的最小花费.
那么转移就是: dp[to][(j+quan)%mod]=min(dp[now][j]+quan)
因为具有后效性,所以需要spfa.
//这道题较多的参考了晚上的解法,在看懂之后自己又写了一遍
//链接:https://blog.csdn.net/qq_33229466/article/details/77131289
#include<cstdio>
#include<queue>
#include<string.h>
using namespace std;
typedef long long LL;
const int maxn = 50 + 5, maxm = 2e4 + 5;
int n, m, num, head[maxn];
LL t, dp[maxn][maxm], inf = 2000000000000000000;
bool vis[maxn][maxm];
struct node {
int to, w, next;
}edge[maxn * 2];
queue<pair<int, int>>q;
void add_edge(int u, int v, int w)
{
edge[++num].to = v;
edge[num].w = w;
edge[num].next = head[u];
head[u] = num;
}
void spfa(int mod)
{
q.push(make_pair(1,0)); vis[1][0] = 1;
while (!q.empty())
{
pair<int,int> u = q.front(); q.pop();
int x = u.first, y = u.second;
for (int i = head[x]; i; i = edge[i].next)
if (dp[x][y] + edge[i].w < dp[edge[i].to][(y + edge[i].w) % mod])
{
int nx = edge[i].to, ny = (y + edge[i].w) % mod;
dp[nx][ny] = dp[x][y] + edge[i].w;
if (!vis[nx][ny]) {
q.push(make_pair(nx, ny));
vis[nx][ny] = 1;
}
}
vis[x][y] = 0;
}
}
int main()
{
int kase;
scanf("%d", &kase);
while (kase--)
{
memset(head, 0, sizeof(head));
scanf("%d%d%lld", &n, &m, &t);
num = 0;
for (int i = 1; i <= m; i++)
{
int x, y, w;
scanf("%d%d%d", &x, &y, &w);
add_edge(x, y, w);
add_edge(y, x, w);
}
int flag = 0;
for (int i = 1; i <= num; i += 2)
if (edge[i].to == 1 || edge[i + 1].to == 1)
{
int w = edge[i].w * 2;
for (int j = 1; j <= n; j++) {
for (int k = 0; k < w; k++) {
dp[j][k] = inf;
}
}
dp[1][0] = 0;
spfa(w);
if (dp[n][t%w] <= t)
{
printf("Yes\n");
flag = 1;
break;
}
}
if (!flag) printf("No\n");
}
return 0;
}