Codeforces Round #257 (Div. 1) (Codeforces 449B)

题意：给力一张无向图，有一些边是正常道路，有一些边是铁路，问最多能删除几条铁路使得所有点到首都（编号为1）的最短路长度不变。

思路：求不能删除的铁路数，总数减掉就是答案。先求出首都到所有点的最短路，求完最短路后，枚举除首都外所有点，如果这个点被更新的边中只有铁路，那么就有一条铁路不能删除。

注意：这里求最短路一开始用SPFA在第45个点TLE，最后换成带堆优化Dijkstra

#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include <iostream>
#include<queue>
#define M 1000010
#define N 100050
#define LL __int64
#define INF (1000000000000000LL)
using namespace std;
struct node {
    int to, nx, va, flag;
} e[M];
int head[N], ecnt;
struct info {
    int id, va;
};
bool operator<(info a, info b) {
    return a.va > b.va;
}
priority_queue<info> q;
void addedge(int x, int y, int va, int flag) {
    e[ecnt].to = y;
    e[ecnt].va = va;
    e[ecnt].nx = head[x];
    e[ecnt].flag = flag;
    head[x] = ecnt++;
}
bool bo[N];
LL d[N];
int n, m, k;
void Dij() {
    for (int i = 0; i <= n; ++i)
        d[i] = INF;
    memset(bo, 0, sizeof(bo));
    d[1] = 0;
    int st = 0, ed = 0;
    while (!q.empty())
        q.pop();
    info a;
    a.va = 0;
    a.id = 1;
    q.push(a);
    while (!q.empty()) {
        a = q.top();
        q.pop();
        int now = a.id;
        if (bo[now])
            continue;
        bo[now] = true;
        for (int j = head[now]; j != -1; j = e[j].nx) {
            int u = e[j].to;
            if (d[u] > d[now] + e[j].va) {
                d[u] = d[now] + e[j].va;
                a.id=u;
                a.va=d[u];
                q.push(a);
            }
        }
    }
}
void init() {
    scanf("%d%d%d", &n, &m, &k);
    memset(head, -1, sizeof(head));
    ecnt = 0;
    for (int i = 0; i < m; ++i) {
        int x, y, z;
        scanf("%d%d%d", &x, &y, &z);
        addedge(x, y, z, 0);
        addedge(y, x, z, 0);
    }
    for (int i = 0; i < k; ++i) {
        int x, y;
        scanf("%d%d", &x, &y);
        addedge(1, x, y, 1);
        addedge(x, 1, y, 1);
    }
}
void solve() {
    Dij();
    int ans = k;
    for (int i = 2; i <= n; ++i) {
        int flag = 0, flag1 = 0;
        for (int j = head[i]; j != -1; j = e[j].nx) {
            int u = e[j].to;
            if (d[u] + e[j].va == d[i]) {
                if (e[j].flag == 1)
                    flag = 1;
                else
                    flag1 = 1;
            }
        }
        if (flag == 1 && flag1 == 0)
            ans--;
    }
    printf("%d
", ans);
}
int main() {
    init();
    solve();

}