Codeforces 449b jzzhu and cities (Shortest Path)

Link: codeforces 449b jzzhu and cities

Jzzhu is the president of a country. There are n cities in this country, with 1 as the capital. m roads already exist and M roads are given. There are also K railroad tracks with the capital and Si distance being Yi. Now, jzzhu wants to save money and dismantle some rails. He asks how many rails can be demolished at most, and the shortest distance between each city and the capital remains unchanged.

Solution: the shortest path: add an array of tags, and mark 1 for each Si. If the shortest path of these points is updated, the mark of the corresponding Si is cleared to 0, finally, count the remaining marks, that is, tracks that cannot be removed.

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;typedef long long ll;typedef pair<int, int> pii;const int maxn = 1e5 + 5;const ll INF = 0x3f3f3f3f3f3f3f3f;int N, M, K, p[maxn], vis[maxn];ll d[maxn];vector<pii> g[maxn];void init () {    scanf("%d%d%d", &N, &M, &K);    int u, v, x;    for (int i = 0; i < M; i++) {        scanf("%d%d%d", &u, &v, &x);        g[u].push_back(make_pair(v, x));        g[v].push_back(make_pair(u, x));    }}int solve () {    for (int i = 0; i <= N; i++)        d[i] = INF;    d[1] = 0;    memset(p, 0, sizeof(p));    queue<int> que;    vis[1] = 1;    que.push(1);    for (int i = 1; i <= K; i++) {        int u, x;        scanf("%d%d", &u, &x);        if (d[u] > x) {            d[u] = x; p[u] = 1;            if (vis[u] == 0) {                vis[u] = 1;                que.push(u);            }        }    }    while (!que.empty()) {        int u = que.front();        que.pop();        vis[u] = 0;        for (int i = 0; i < g[u].size(); i++) {            int v = g[u][i].first, x = g[u][i].second;            if (d[v] >= d[u] + x && p[v])                p[v] = 0;            if (d[v] > d[u] + x) {                d[v] = d[u] + x;                if (vis[v] == 0) {                    vis[v] = 1;                    que.push(v);                }            }        }    }    for (int i = 1; i <= N; i++)        K -= p[i];    return K;}int main () {    init();    printf("%d\n", solve());    return 0;}

Codeforces 449b jzzhu and cities (Shortest Path)