2013 腾讯马拉松初赛第二场

　　转换成秒后模拟即可

View Code

#include<stdio.h>
#include<stdlib.h>

int main(){
    int T;
    scanf("%d", &T);
    while( T-- ){
        int h1,m1,s1;
        int h2,m2,s2;
        int mod = 12*60*60;    
        scanf("%d:%d:%d",&h1,&m1,&s1);
        scanf("%d:%d:%d",&h2,&m2,&s2);
        s1 = h1*3600+m1*60+s1;
        s2 = (h2*3600+m2*60+s2)%mod;
        
        s1 = (s1-s2+mod)%mod;
        h2 = s1/3600; m2 = (s1%3600)/60; s2 = s1%60;    
        printf("%02d:%02d:%02d\n", h2, m2, s2 );

    }
    return 0;
}

B 小明系列故事——女友的考验

　　据说是 AC自动机，但是不会

C 吉哥系列故事——完美队形I

　　反置，对两串求最长上升公共子序列，注意中间下标问题

View Code

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <cmath>
using namespace std;

int seq[205], rseq[205], N;
int f[205][205];

inline int Min(int a, int b) {
    return a > b ? b : a;
}

inline int Max(int a, int b) {
    return a > b ? a : b;
}

void solve(int a[], int b[]) {
    int max;
    int Len = -1;
    int n1 = N, n2 = N;
    for(int i=1;i<=n1;i++)
    {
        max=0;
        for(int j=1;j<=n2;j++)
        {
            f[i][j]=f[i-1][j];
            if (a[i]>b[j]&&max<f[i-1][j]) {
                max=f[i-1][j];
            }
            if (a[i]==b[j]) {
                f[i][j]=max+1;
                if (N-j+1 > i) {
                    Len = Max(Len, f[i][j]*2);
                }
                else if (N-j+1 == i){
                    Len = Max(Len, f[i][j]*2-1);
                }
            }
        }
    }
    printf("%d\n", Min(Len, N));
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &N);
        for (int i = 1; i <= N; ++i) {
            scanf("%d", seq+i);
            rseq[N-i+1] = seq[i];
        }
        solve(seq, rseq);
    }
    return 0;
}

D 吉哥系列故事——完美队形II

　　分治+扩展KMP，或者另外一种处理字符串 M啥

E 湫湫系列故事——设计风景线

　　并查集+点分治，往上更新的时候合并子树，这样就不会超时

View Code

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAX(a,b) (a)>(b)?(a):(b)
const int N = 1e5+10;
const int M = 1e6+10;

int n, m;

struct node{
    int v, c, nxt;
}edge[M<<2];
struct Edge{
    int u, v, c;
    void input(){
        scanf("%d%d%d",&u,&v,&c);    
        u--; v--;    
    }
}Q[M];

int head[N], idx;
int st[N], dis[N];
int res;
bool vis[N];

int find( int x ){
    return (x==st[x]) ? x : (st[x]=find(st[x])) ;
}
bool legal(){
    
    for(int i = 0; i < n; i++) st[i] = i;
    for(int i = 0; i < m; i++){
        int u = Q[i].u, v = Q[i].v;
        int x = find(u), y = find(v);
        if( x == y ) return false;
        else st[x] = y;
    } 
    return true;
}
void addedge(int u,int v,int c){
    edge[idx].v = v; edge[idx].c = c;
    edge[idx].nxt = head[u]; head[u] = idx++;

    edge[idx].v = u; edge[idx].c = c;
    edge[idx].nxt = head[v]; head[v] = idx++;
}
void dfs(int u,int pre ){
    if( !vis[u] ) vis[u] = true;

    for( int i = head[u]; ~i; i = edge[i].nxt ){    
        int v = edge[i].v;
        //printf("v = %d\n", v );    
        if( v != pre ){    
            dfs( v, u );
            res = MAX( res, dis[u]+dis[v]+edge[i].c );

            dis[u] = MAX( dis[u], dis[v]+edge[i].c );
        }
    } 
}

int solve(){
    memset( head, 0xff, sizeof(head));
    idx = 0;
    for(int i = 0; i < m; i++)
        addedge( Q[i].u, Q[i].v, Q[i].c );    
    
    res = 0;
    memset( vis, 0, sizeof(vis));
    memset( dis, 0, sizeof(dis));
    for(int i = 0; i < n; i++)
        if( !vis[i] ) dfs( i, -1 );
    return res;
}
int main(){
    while( scanf("%d%d",&n,&m) != EOF)
    {
        for(int i = 0; i < m; i++)
            Q[i].input();
        if( !legal() ) puts("YES");
        else    printf("%d\n", solve() );
    }
    return 0;
}