cf567E. President and Roads(最短路计数)

题意

题目链接

给出一张有向图，以及起点终点，判断每条边的状态：

1. 是否一定在最短路上，是的话输出'yes'

2. 如果不在最短路上，最少减去多少权值会使其在最短路上，如果减去后的权值\(< 1\)，输出'no'，否则输出'can + 花费'

sol

考察对最短路的理解。

首先确定哪些边一定在最短路上，一个条件是 从起点到该点的最短路 + 边权 + 从该点到终点的最短路 = 从起点到终点的最短路

同时还要满足没有别的边可以代替这条边，可以用tarjan求一下桥。当然也可以直接用最短路条数判

这样的话正反跑一边dijkstra求出最短路以及最短路径的条数，判断一下即可

#include<bits/stdc++.h>
#define pair pair<ll, int> 
#define mp make_pair
#define fi first
#define se second 
#define ll  long long
using namespace std;
const int maxn = 2e5 + 10;
const ll inf = 1e18 + 10; 
const ll mod1 = 2860486313ll, mod2 = 1500450271ll; 
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, s, t, vis[maxn];
ll f[maxn], g[maxn], f2[maxn], g2[maxn];
ll dis[maxn], rdis[maxn];
vector<pair> v[maxn], t[maxn];
struct edge { int u, v; ll w;} e[maxn];
ll add(ll x, ll y, ll mod) {
    return (x + y >= mod ? x + y - mod : x + y);
}
void dij(int s, ll *d, ll *f, ll *f2, int opt) {
    priority_queue<pair> q; q.push(mp(0, s));
    for(int i = 1; i <= n; i++) d[i] = inf;
    d[s] = 0; f[s] = f2[s] = 1; memset(vis, 0, sizeof(vis)); 
    while(!q.empty()) {
        if(vis[q.top().se]) {q.pop(); continue;}
        int p = q.top().se; q.pop(); vis[p] = 1;
        vector<pair> *e = (opt == 1 ? v + p : t + p);
        for(int i = 0; i < e -> size(); i++) {
            int to = (*e)[i].fi, w = (*e)[i].se;
            if(d[to] > d[p] + w) d[to] = d[p] + w, f[to] = f[p], f2[to] = f2[p], q.push(mp(-d[to], to));
            else if(d[to] == d[p] + w) f[to] = add(f[to], f[p], mod1), f2[to] = add(f2[to], f2[p], mod2);
        }
    }
}
signed main() {
    n = read(); m = read(); s = read(); t = read();
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read(), z = read(); e[i] = (edge) {x, y, z};
        v[x].push_back(mp(y, z)); 
        t[y].push_back(mp(x, z));
    }
    dij(s, dis, f, f2, 1); 
    dij(t, rdis, g, g2, 2);
    for(int i = 1; i <= m; i++) {
        int x = e[i].u, y = e[i].v;ll w = e[i].w;
        if((dis[x] + w + rdis[y] == dis[t]) && (1ll * f[x] * g[y] % mod1 == f[t]) && (1ll * f2[x] * g2[y] % mod2 == f2[t])) puts("yes");
        else {
            ll ned = dis[t] - dis[x] - rdis[y] ;
            if(ned <= 1) puts("no");
            else printf("can %i64d\n", w - ned + 1);
        }
    }
    return 0;
}