Xeon second scrimmage of training camp by Los valley ACM

A. explosion problem in graph theory, shortest path, thinking

Gives a view from 1:00 to 1 meter per second, the start speed of flame propagation, is greater than if they encounter a flame will explode, the final explosion of the number of Q

Easy to draw the following rules: if the degree of explosion at some point, then the point is that the presence of greater than 1 shortest equal   if the explosion at the edge, then this edge is not the shortest path

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
const int maxn = 300000 + 10;
const double PI = acos(-1.0);
typedef long Long LL;
 the using  namespace STD; 

int n-, m; 

struct Edge { int V, w; Edge ( int A, int B) {V = A, w = B;}}; // V end point, w right side 

struct Node {
     int ID, dis; // ID dis point number temporal distance 
    Node ( int A, int B) {A = ID, dis = B;} 
    Friend BOOL  operator < (Node A, Node B) {
         return a.dis> B .dis; // each team let out a small distance 
    } 
}; 

the Vector <Edge>E [MAXN];
 int DIS [MAXN]; // record the shortest 
BOOL VIS [MAXN]; // record is found shortest 
int ANS; 

void the Dijkstra () {
     int S = . 1 ; // S under the circumstances to change the starting point 
    for ( int I = 0 ; I <= n-; I ++) DIS [I] = INF, VIS [I] = 0 ; 
    DIS [S] = 0 ; 
    The priority_queue <Node> Q; 
    Q.push (Node (S, 0 ) );
     the while (! ) {Q.empty () 
        Node U = Q.top (); Q.pop ();
        if (vis[u.id])continue;
        vis[u.id] = 1;
        for (int i = 0; i < e[u.id].size(); i++) {//遍历邻居
            edge y = e[u.id][i];
            if (vis[y.v])continue;
            if (dis[y.v] > y.w + dis[u.id]) {
                dis[y.v] = y.w + u.dis;
                Q.push(node(y.v, dis[y.v]));//更新最短路
            }
        }
    }

}
int main() {
    scanf("%d%d", &n, &m);
    int u, v, w;
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &u, &v, &w);
        e[u].push_back(edge(v, w));
        if (u != v) e[v].push_back(edge(u, w));
    }
    Dijkstra();
    //for (int i = 1; i <= n; i++) printf("%d  ", dis[i]);
    for (int i = 1; i <= n; i++) {
        int cnt = 0;
        for (int j = 0; j < e[i].size(); j++) {
            if (dis[i] == dis[e[i][j].v] + e[i][j].w) cnt++;
            else if (i >= e[i][j].v && dis[i] + e[i][j].w > dis[e[i][j].v] && dis[e[i][j].v] + e[i][j].w > dis[i]) ans++;  //core
        }
        if (cnt > 1) ans++;
    }
    printf("%d", ans);
    return 0;
}

View Code

 

B. simulation title / title formula

  In fact, gives the date, begged date n days

#include <bits/stdc++.h>
using namespace std;
inline void read(int &a) {
    int ch = getchar();
    while (ch < '0') ch = getchar();
    a = 0;
    while (ch >= '0') a = a * 10 + ch - '0', ch = getchar();
}
int M[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
void work() {
    int Mo, y, m, d;
    read(Mo); read(y); read(m); read(d); 
    int del = 0;
    while (Mo < int(1e9)) {
        del++;
        Mo += del;
        if (++d > M[m] + (m == 2 && y % 4 == 0 && (y % 100 != 0 || y % 400 == 0))) {
            d = 1;
            if (++m == 13) y ++, m = 1;
        }
    }
    printf("%d %d %d\n", y, m, d);
}

int main() {
    int T; read(T);
    while (T--) work();
    return 0;
}

View Code

 

C. next time

D. given a, b, k, gcd (a, b) = 1 find satisfy the equation ax + by = c, ab k-th small c <0 is