首先这个题目问的是从\(1\)到\(n\)的路径问题.所以我们可以先将与这些路径无关的边权直接赋为你喜欢的数\(\in\{1,2\}\).
令\(dis[i]\)表示确定玩\(w[i]\)后从\(1\)到\(i\)的最短距离.
那么对于剩下的任意一条边\(\{u,v\}\),必定有\(1\leq dis[v]-dis[u]\leq2\).
其实也很好理解,就是\(1\leq w[i]\leq2\)并且每一条路径长度要相同.
如果\(dis[v]-dis[u]>2\)且\(dis[v]-dis[u]<1\),就一定出现了路径不同的边(\(w[i]\)无法弥补他们的差距).
于是我们就有了若干个约束.
于是将\(u\rightarrow v\)连一条\(1\)的边,将\(v\rightarrow u\)连一条\(-2\)的边.
跑差分约束即可.
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define il inline
#define rg register
#define gi read<int>
#define pii pair<int, int>
using namespace std;
const int N = 1010, M = 5010;
template<class TT>
il TT read() {
TT o = 0,fl = 1; char ch = getchar();
while (!isdigit(ch) && ch != '-') ch = getchar();
if (ch == '-') fl = -1, ch = getchar();
while (isdigit(ch)) o = o * 10 + ch - '0', ch = getchar();
return fl * o;
}
queue<int>Q;
pii book[M];
int mp[N][N], n, m, dis[N], cnt[N];
bool vis[2][N], mark;
il void dfs(int u) {
vis[mark][u] = 1;
for (int v = 1; v <= n; ++v)
if (mp[u][v] && !vis[mark][v])
vis[mark][v] = 1, dfs(v);
}
il void spfa() {
memset(dis, -63, sizeof dis);
Q.push(1); cnt[1] = 1; dis[1] = 0;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int v = 1; v <= n; ++v)
if (mp[u][v] && dis[v] < dis[u] + mp[u][v]) {
dis[v] = dis[u] + mp[u][v];
Q.push(v); cnt[v]++;
if (cnt[v] > n) exit(puts("No") & 0);
}
}
}
int main() {
n = gi(), m = gi();
for (int i = 1; i <= m; ++i) {
int u = gi(), v = gi();
book[i] = pii(u, v);
mp[u][v] = 1;
}
mark = 0; dfs(1);
for (int i = 1; i <= m; ++i) {
mp[book[i].first][book[i].second] = 0;
mp[book[i].second][book[i].first] = 1;
}
mark = 1; dfs(n);
for (int i = 1; i <= m; ++i) {
int u = book[i].first, v = book[i].second;
mp[v][u] = 0;
if (vis[0][u] && vis[1][u] && vis[0][v] && vis[1][v])
mp[u][v] = 1, mp[v][u] = -2;
// else printf("%d\n", i);
}
spfa(); puts("Yes");
for (int i = 1; i <= m; ++i) {
int u = book[i].first, v = book[i].second;
if (mp[u][v]) printf("%d\n", dis[v] - dis[u]);
else puts("2");
}
return 0;
}