2013 腾讯马拉松初赛第二场

A 小Q系列故事——为什么时光不能倒流

　　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;

}