Codeforces 567E President and Roads 最短路 + tarjan求桥

很套路的题啊，见过很多次了。
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, m, s, t;
int a[N], b[N], l[N], cnt[N], ans[N];
LL d[2][N];
vector<vector<PLI>> G, rG, tG;

void dij(int s, LL d[N], vector<vector<PLI>> &G) {
    priority_queue<PLI, vector<PLI>, greater<PLI> > que;
    d[s] = 0; que.push(mk(0, s));
    while(!que.empty()) {
        int u = que.top().se;
        LL dis = que.top().fi;
        que.pop();
        if(dis > d[u]) continue;
        for(auto &e : G[u]) {
            if(chkmin(d[e.se], dis + e.fi)) {
                que.push(mk(d[e.se], e.se));
            }
        }
    }
}

int dfn[N], low[N], idx;

void tarjan(int u, int id) {
    dfn[u] = low[u] = ++idx;
    for(auto &e : tG[u]) {
        if(e.fi == id) continue;
        if(!dfn[e.se]) {
            tarjan(e.se, e.fi);
            low[u] = min(low[u], low[e.se]);
            if(dfn[u] < low[e.se]) ans[e.fi] = 0;
        } else low[u] = min(low[u], dfn[e.se]);
    }
}

int main() {
    memset(d, 0x3f, sizeof(d));
    scanf("%d%d%d%d", &n, &m, &s, &t);
    G.resize(n + 1); rG.resize(n + 1); tG.resize(n + 1);
    for(int i = 1; i <= m; i++) {
        scanf("%d%d%d", &a[i], &b[i], &l[i]);
        G[a[i]].push_back(mk(l[i], b[i]));
        rG[b[i]].push_back(mk(l[i], a[i]));
    }
    dij(s, d[0], G);
    dij(t, d[1], rG);
    for(int i = 1; i <= m; i++) {
        if(d[0][a[i]] + d[1][b[i]] + l[i] >= INF) {
            ans[i] = -1;
        } else {
            if(d[0][a[i]] + d[1][b[i]] + l[i] == d[0][t]) {
                tG[a[i]].push_back(mk(i, b[i]));
                tG[b[i]].push_back(mk(i, a[i]));
            }
            ans[i] = d[0][a[i]] + d[1][b[i]] + l[i] + 1 - d[0][t];
        }
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i]) {
            tarjan(i, 0);
        }
    }
    for(int i = 1; i <= m; i++) {
        if(ans[i] == -1 || l[i] - ans[i] <= 0) puts("NO");
        else if(ans[i] == 0) puts("YES");
        else printf("CAN %d
", ans[i]);
    }
    return 0;
}

/*
*/