2014 ACM/ICPC Asia Regional Xi'an Online

A Post Robot

。。。嗯，比赛时手拍了一个自动机，注意是手拍。。= =

 1 /*
 2 ID:esxgx1
 3 LANG:C++
 4 PROG:A
 5 */
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 
12 char str[10007];
13 
14 int main(void)
15 {
16     #ifndef ONLINE_JUDGE
17     freopen("in.txt", "r", stdin);
18     #endif
19     while(gets(str) != NULL) {
20         int dd = 0;
21         for(int i = 0; str[i]; ++i) {
22 _failed:
23             //printf("%c %d\n", str[i], dd);
24             if (dd == 0) {
25                 if (str[i] == 'A') dd = 1;
26                 else if (str[i] == 'i') dd = 5;
27                 else if (str[i] == 'S') dd = 11;
28             }
29             else if (dd == 1) {
30                     if (str[i] == 'p') dd = 2;
31                     else {dd = 0; goto _failed;}
32             }
33             else if (dd == 2) {
34                     if (str[i] == 'p') dd = 3;
35                     else {dd = 0; goto _failed;}
36             }
37             else if (dd == 3) {
38                     if (str[i] == 'l') dd = 4;
39                     else {dd = 0; goto _failed;}
40             }
41             else if (dd == 4) {
42                     dd = 0;
43                     if (str[i] == 'e') printf("MAI MAI MAI!\n");
44                     else {dd = 0; goto _failed;}
45             }
46             else if (dd == 5) {
47                 if (str[i] == 'P') dd = 6;
48                 else {dd = 0; goto _failed;}
49             }
50             else if (dd == 6) {
51                 if (str[i] == 'h') dd = 7;
52                 else if (str[i] == 'a') dd = 10;
53                 else if (str[i] == 'o') dd = 10;
54                 else {dd = 0; goto _failed;}
55             }
56             else if (dd == 7) {
57                 if (str[i] == 'o') dd = 8;
58                 else {dd = 0; goto _failed;}
59             }
60             else if (dd == 8) {
61                 if (str[i] == 'n') dd = 9;
62                 else {dd = 0; goto _failed;}
63             }
64             else if (dd == 9) {
65                 dd = 0;
66                 if (str[i] == 'e') printf("MAI MAI MAI!\n");
67                 else {dd = 0; goto _failed;}
68             }
69             else if (dd == 10) {
70                 dd = 0;
71                 if (str[i] == 'd') printf("MAI MAI MAI!\n");
72                 else {dd = 0; goto _failed;}
73             }else if (dd == 11) {
74                 if (str[i] == 'o') dd = 12;
75                 else {dd = 0; goto _failed;}
76             } else if (dd == 12) {
77                 if (str[i] == 'n') dd = 13;
78                 else {dd = 0; goto _failed;}
79             } else if (dd == 13) {
80                 dd = 0;
81                 if (str[i] == 'y') printf("SONY DAFA IS GOOD!\n");
82                 else {dd = 0; goto _failed;}
83             }
84         }
85     }
86     return 0;
87 }

 

C Paint Pearls

状态方程好想，但预处理真的不是很容易。简单来说预处理用到了一个单调性，即颜色不同的越多越要往回看。

具体分析干脆引用某大神的：

状态转移方程很好想，dp[i] = min(dp[j]+o[j~i]^2,dp[i]) ,o[j~i]表示从j到i颜色的种数。
普通的O（n*n）是会超时的，可以想到o[]最大为sqrt(n),问题是怎么快速找到从i开始往前2种颜色、三种、四种。。。o[]种的位置。
离散化之后，可以边走边记录某个数最后一个出现的位置，初始为-1，而所要求的位置就等于
if(last[a[i]]==-1) 该数没有出现过，num[i][1] = i,num[i][j+1] = num[i-1][j];
else  last[a[i]]之前 num[i][1] = i,num[i][j+1] = num[i-1][j],之后num[i][j]= num[i-1][j];

当然赛时我这个渣渣是肯定写不出来的，后来看了这个分析写了一份代码，个人感觉按这个思路真的不是很好写（也许应该看看thu的那份）：

/*
ID:esxgx1
LANG:C++
PROG:xian_C
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define NN        500007

int coloridx[NN], color[NN], ncolors;

int cpre[NN];                // 某个颜色上一次出现的位置
int clbck[NN];                // n颜色回溯

int _dp[NN+1], *dp = &_dp[1];

int main(void)
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    #endif

    int N;
    while(scanf("%d", &N) > 0) {
        for(int i=0; i<N; ++i) {
            scanf("%d", &color[i]);
            coloridx[i] = color[i];
        }
        // 颜色离散化
        sort(color, color + N);
        ncolors = unique(color, color + N) - color;

        for(int i=0; i<N; ++i)
            coloridx[i] = lower_bound(color, color + ncolors, coloridx[i]) - color;


        memset(clbck, -1, sizeof(clbck));
        memset(cpre, -1, sizeof(cpre));

        clbck[1] = cpre[coloridx[0]] = 0;
        dp[0] = 1;

        for(int i=1; i<N; ++i) {
            dp[i] = 0x3f3f3f3f;
            if (cpre[coloridx[i]] != i-1) {
                if (cpre[coloridx[i]] < 0)    {    // 颜色没出现过
                    for(int j = sqrt(N); j>0; --j) clbck[j+1] = clbck[j];
                } else {
                    int j = sqrt(N) + 1;
                    for(; j > 0 && (clbck[j] < 0 || clbck[j] <= cpre[coloridx[i]]); --j);
                    for(--j; j > 0 && clbck[j] > cpre[coloridx[i]]; --j) clbck[j+1] = clbck[j];
                }
                clbck[1] = i;
            }
            for(int j = 1; clbck[j] >= 0; ++j) {
                dp[i] = min(dp[i], dp[clbck[j]-1] + j*j);
                //printf("(%d) j=%d %d [pre%d]\n", i+1, j, clbck[j]+1, cpre[coloridx[i]] + 1);
            }

            //printf("%d %d\n", i, dp[i]);
            cpre[coloridx[i]] = i;
        }
        printf("%d\n", dp[N-1]);

    }
    return 0;
}

 

E Game

真的很凑巧，这题是我们的省赛原题，然后又yy了一份代码。。

#include <cstdio>
using namespace std;
 
int N;
 
int main()
{
    while(scanf("%d", &N) > 0) {
        int i, j;
        j = 0;
        for(i=0; i<N; ++i) {
            int k;
            scanf("%d", &k);
            j ^= k;
        }
        printf(j? "Win\n" : "Lose\n");
    }
    return 0;
}

 F Dice

这题我没参与，不知道讲了神马。。听队友说是BFS判重，贴一下我们队的代码。。

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<queue>
 5 using namespace std;
 6 int hash[700000];
 7 int dis[700000];
 8 struct dice1{
 9     int top,but,lef,rig,fro,bac;
10 };
11 int enc(int a,int b,int c,int d,int e,int f){
12     return a*100000+b*10000+c*1000+d*100+e*10+f;
13 }
14 int enc(dice1 tmp){
15     return tmp.top*100000+tmp.but*10000+tmp.lef*1000+tmp.rig*100+tmp.fro*10+tmp.bac;
16 }
17 dice1 dice;
18 dice1 rol(dice1 tmp){
19     dice1 ret;
20     ret.top=tmp.rig;
21     ret.rig=tmp.but;
22     ret.but=tmp.lef;
23     ret.lef=tmp.top;
24     ret.fro=tmp.fro;
25     ret.bac=tmp.bac;
26     return ret;
27 }
28 dice1 ror(dice1 tmp){
29     dice1 ret;
30     ret.top=tmp.lef;
31     ret.rig=tmp.top;
32     ret.but=tmp.rig;
33     ret.lef=tmp.but;
34     ret.fro=tmp.fro;
35     ret.bac=tmp.bac;
36     return ret;
37 }
38 dice1 rof(dice1 tmp){
39     dice1 ret;
40     ret.top=tmp.bac;
41     ret.fro=tmp.top;
42     ret.but=tmp.fro;
43     ret.bac=tmp.but;
44     ret.lef=tmp.lef;
45     ret.rig=tmp.rig;    
46     return ret;
47 }
48 dice1 rob(dice1 tmp){
49     dice1 ret;
50     ret.top=tmp.fro;
51     ret.fro=tmp.but;
52     ret.but=tmp.bac;
53     ret.bac=tmp.top;
54     ret.lef=tmp.lef;
55     ret.rig=tmp.rig;
56     return ret;
57 }
58 int main(){
59     int a[7],b[7];
60     while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])>0){
61         memset(hash,0,sizeof(hash));
62         memset(dis,0,sizeof(dis));
63         scanf("%d%d%d%d%d%d",&b[1],&b[2],&b[3],&b[4],&b[5],&b[6]);
64         hash[enc(b[1],b[2],b[3],b[4],b[5],b[6])]=2;
65         queue<dice1> q;
66         dice1 aaa;
67         aaa.top=a[1];
68         aaa.but=a[2];
69         aaa.lef=a[3];
70         aaa.rig=a[4];
71         aaa.fro=a[5];
72         aaa.bac=a[6];
73         q.push(aaa);
74         dis[enc(q.front())]=0;
75         int ans=-1;
76         while(!q.empty()){
77             dice1 tmp=q.front();q.pop();
78             if(hash[enc(tmp)]==2){
79                 ans=dis[enc(tmp)];
80                 break;
81             }
82             if(dis[enc(tmp)]>4){
83                 ans=-1;break;
84             }
85             if(hash[enc(tmp)]!=1){
86                 dis[enc(rol(tmp))]=dis[enc(tmp)]+1;
87                 q.push(rol(tmp));
88                 dis[enc(ror(tmp))]=dis[enc(tmp)]+1;
89                 q.push(ror(tmp));
90                 dis[enc(rof(tmp))]=dis[enc(tmp)]+1;
91                 q.push(rof(tmp));
92                 dis[enc(rob(tmp))]=dis[enc(tmp)]+1;
93                 q.push(rob(tmp));
94             }
95         }
96         printf("%d\n",ans);
97     }
98     return 0;
99 }

 

H Number Sequence

#include<cstdio>
#include<cstring>
using namespace std;

int n;
long long ans[100007];

int a[100007];

int main()
{
      int i;
      long long x,sum;
      while (scanf("%d",&n) > 0) {
             for(int i=0; i<=n; ++i)
                scanf("%d", &a[i]);
             x = 1;
             while (x<=n) x <<= 1;
             x--;
             memset(ans,-1,sizeof(ans));

             for (i=n; i>=0; --i) {
                   if (ans[i]!=-1) continue;
                   while ((x^i)>n || ans[x^i]!=-1) x>>=1;
                   ans[x^i]=i,ans[i]=x^i;
             }
             sum=0;
             for (i=0; i<=n; ++i)
                   sum+=i^ans[i];
             printf("%I64d\n",sum);
             for (i=0;i<=n;i++) {
                if (i) putchar(' ');
                printf("%I64d",ans[a[i]]);
             }
             putchar('\n');
      }
      return 0;
}

 

I 233 Matrix

快速幂，神队友脑洞大开yy了这个矩阵，然后就没有然后了。。

 1 /*
 2 ID:esxgx1
 3 LANG:C++
 4 PROG:I
 5 */
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 
12 
13 #define MM 17
14 
15 struct matrix {
16     long long m[MM][MM];
17 };
18 
19 #define MOD        10000007
20 
21 int Nn;
22 
23 struct matrix m_mult(const struct matrix &a, const struct matrix &b)
24 {
25     struct matrix r;
26     for(int i=0; i<Nn; ++i) {
27         for(int j=0; j<Nn; ++j) {
28             r.m[i][j] = 0;
29             for(int k = 0; k < Nn; ++k) {
30                 r.m[i][j] += a.m[i][k] * b.m[k][j];
31                 r.m[i][j] %= MOD;
32             }
33         }
34     }
35     return r;
36 }
37 
38 struct matrix m_pow(struct matrix a, int n)
39 {
40     struct matrix p;
41     memset(&p, 0, sizeof(p));
42     for(int i=0; i<Nn; ++i)
43         p.m[i][i] = 1;
44     while(n) {
45         if (n & 1) p = m_mult(p, a);
46         a = m_mult(a, a);
47         n >>= 1;
48     }
49     return p;
50 }
51 
52 int main(void)
53 {
54     #ifndef ONLINE_JUDGE
55     freopen("in.txt", "r", stdin);
56     #endif
57 
58     int N, M;
59     while(scanf("%d%d", &N, &M) > 0) {
60         ++N;
61         matrix p;
62         memset(&p, 0, sizeof(p));
63         for(int i=0; i < N; ++i) {
64             p.m[i][0] = 10;
65             for(int j=1; j<=i; ++j)
66                 p.m[i][j] = 1;
67             p.m[i][N] = 1;
68         }
69         p.m[N][N] = 1, Nn = N+1;
70         p = m_pow(p, M);
71 
72         matrix q;
73         memset(&q, 0, sizeof(q));
74         q.m[0][0] = 23, q.m[N][0] = 3;
75         for(int i=1; i<N; ++i)
76             scanf("%I64d", &q.m[i][0]);
77         p = m_mult(p, q);
78         //p = q;
79         printf("%d\n", p.m[N-1][0]);
80     }
81     return 0;
82 }

 

posted @ 2014-09-15 02:34  e0e1e  阅读(369)  评论(0)  编辑  收藏  举报