题目地址:
http://acdream.info/onecontest/1052#overview
A:
简单的字符串问题
1 // File Name: a.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月11日 星期五 19时06分11秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 #include<climits> 25 #include<queue> 26 27 using namespace std; 28 char str[200]; 29 int main(){ 30 int t ; 31 scanf("%d",&t); 32 char str1[120]; 33 while(t--) 34 { 35 memset(str,0,sizeof(str)); 36 memset(str1,0,sizeof(str1)); 37 scanf("%s",str); 38 for(int i = 0 ;; i ++ ) 39 { 40 str1[i] = 'A'; 41 if(!strstr(str,str1)) 42 { 43 puts(str1); 44 break; 45 } 46 } 47 } 48 49 return 0; 50 }
B:
搜索DP博弈
1 // File Name: b.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月12日 星期六 10时44分56秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 #include<climits> 25 #include<queue> 26 27 using namespace std; 28 long long n , m , a; 29 int log10(long long n ,long long a) 30 { 31 double t = (log(a)/log(n)); 32 return (t+1.0 - 1e-8); 33 34 } 35 long long P(long long x) 36 { 37 long long ans = 1; 38 for(int i =1 ;i <= x ;i ++) 39 ans *= 2 ; 40 return ans ; 41 } 42 int dfs(long long n,long long m ) 43 { 44 if(n != 1) 45 { 46 double t = (log(a)/log(n)); 47 long long k = (t+1.0 - 1e-8); 48 if(k <= m ) 49 { 50 return 1; 51 } 52 } 53 int t1; 54 int t2; 55 if(m == 1 && n*n >= a ) 56 { 57 t1 = 1; 58 t2 = (a - n )%2; 59 } 60 else if(n == 1 && P(m) >= a) 61 { 62 t1 = 1; 63 t2 = dfs(n+1,m); 64 }else{ 65 t1 = dfs(n+1,m); 66 t2 = dfs(n,m+1); 67 } 68 if(t1 && t2) 69 return 0 ; 70 if(!t1 && !t2) 71 return 1; 72 return 1; 73 } 74 int main(){ 75 // freopen("cin.txt","r",stdin); 76 // freopen("cout1.txt","w",stdout); 77 78 while(scanf("%lld %lld %lld",&n,&m,&a) != EOF) 79 { 80 if(n == 1) 81 { 82 if(log10(n+1,a) <= m) 83 { printf("draw "); 84 continue; 85 } 86 int t1 = dfs(n,m); 87 } 88 if(n!= 1) 89 if(log10(n,a) <= m) 90 { 91 printf("lose "); 92 continue; 93 } 94 int t = dfs(n,m); 95 if(t == 1 ) 96 printf("win "); 97 else printf("lose "); 98 } 99 return 0; 100 }
C:
状态压缩+dp
这个题目状态压缩需要考虑一些情况,
一共有三种情况, 被保护,没有被保护,还有落子点.
然后就是对所有情况进行讨论即可.
1 // File Name: c.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月13日 星期日 16时44分27秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 #include<climits> 25 #include<queue> 26 27 using namespace std; 28 #define di 3 29 int dp[11][60000]; 30 int hs[2000000]; 31 int n,m,row; 32 int num[10],pp[10]; 33 int visit[10][10]; 34 void change1(int zt) 35 { 36 int now = zt; 37 int k = 0 ; 38 int num[20]; 39 memset(num,0,sizeof(num)); 40 while(now) 41 { 42 num[++k] = now % di; 43 now/= di; 44 } 45 for(int i =1 ;i<= 10;i ++) 46 printf("%d ",num[i]); 47 printf(" "); 48 } 49 50 void dfs(int pos ,int status,int a) 51 { 52 if(pos > n ) 53 { 54 // printf("%d %d ",status,a); 55 // change1(status); 56 if(dp[row][status] == -1) 57 dp[row][status] = a; 58 else{ 59 // printf("update %d %d ",dp[row][status],a); 60 dp[row][status] = min(dp[row][status],a); 61 } 62 return ; 63 } 64 if(pos == 1){ 65 if(num[pos] == 1) 66 { 67 dfs(pos+1,status+2*pp[pos],a); 68 }else if(num[pos] == 2){ 69 if(visit[row][pos] == 0) 70 dfs(pos+1,status+pp[pos],a+1); 71 dfs(pos+1,status,a); 72 }else { 73 if(visit[row][pos] == 1) 74 return; 75 else dfs(pos+1,status+pp[pos],a+1); 76 } 77 return ; 78 } 79 int t = status/pp[pos-1]; 80 if(t == 1){ 81 if(num[pos] == 0 ) 82 return; 83 else 84 dfs(pos+1,status + 2*pp[pos],a); 85 }else if(t == 2){ 86 if(num[pos] == 1){ //左边为2 上边为1 87 dfs(pos+1,status+2*pp[pos],a); 88 }else if(num[pos] == 2){ //左边为2 上边为2 , 89 if(visit[row][pos] == 0) 90 { 91 dfs(pos+1,status+pp[pos],a+1); 92 } 93 dfs(pos+1,status,a); 94 }else { 95 if(visit[row][pos] == 0) 96 { 97 dfs(pos+1,status+pp[pos],a+1); 98 }else{ 99 return ; 100 } 101 } 102 }else { 103 if(num[pos] == 1){ //左边为0 上边为1 104 dfs(pos+1,status+ 2*pp[pos],a); 105 }else if(num[pos] == 2){ //左边为0 上边为2 , 106 if(visit[row][pos] == 0) 107 { 108 dfs(pos+1,status+pp[pos]+2*pp[pos-1],a+1); 109 } 110 dfs(pos+1,status,a); 111 }else {//左边为0 上边为0 112 if(visit[row][pos] == 0) 113 { 114 dfs(pos+1,status+pp[pos]+2*pp[pos-1],a+1); 115 }else{ 116 return ; 117 } 118 } 119 } 120 121 } 122 void change(int zt) 123 { 124 int now = zt; 125 int k = 0 ; 126 memset(num,0,sizeof(num)); 127 if(now == 0 ) 128 return ; 129 while(now){ 130 num[++k] = now % di; 131 now/= di; 132 } 133 } 134 int ok(int zt) 135 { 136 //printf("%d** ",zt); 137 for(int k = n; k >= 1 ; k -- ) 138 { 139 if(zt/pp[k] == 0 ) 140 return 0 ; 141 zt = zt % pp[k]; 142 } 143 return 1; 144 } 145 int main(){ 146 pp[0] = 1; 147 pp[1] = 1; 148 for(int i = 2;i <= 10 ;i ++) pp[i] = pp[i-1]*di; 149 while(scanf("%d %d",&n,&m) != EOF) 150 { 151 152 memset(visit,0,sizeof(visit)); 153 memset(num,0,sizeof(num)); 154 for(int i =1;i <= m; i ++) 155 { 156 int a,b; 157 scanf("%d %d",&a,&b); 158 visit[a][b] = 1; 159 } 160 memset(dp,-1,sizeof(dp)); 161 int temp = 0 ; 162 for(int i =1; i <= n;i ++) 163 temp += 2* pp[i]; 164 dp[0][temp] = 0 ; 165 166 for(row = 1 ;row <= n;row ++) 167 { 168 for(int i = 0;i <= pp[n+1]; i ++) 169 { 170 if(dp[row-1][i] != -1) 171 { 172 change(i); 173 dfs(1,0,dp[row-1][i]); 174 } 175 } 176 // printf("***** "); 177 } 178 int ans = 1000 ; 179 for(int i = 0 ;i <= pp[n+1]; i ++) 180 { 181 if(dp[n][i] != -1) 182 { 183 if(dp[n][i] < ans && ok(i)) 184 { 185 ans = min(dp[n][i],ans); 186 } 187 } 188 } 189 if(ans == 1000) 190 printf("-1 "); 191 else 192 printf("%d ",ans); 193 } 194 return 0; 195 }
F:
kruskal + 贪心
1 // File Name: f.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月11日 星期五 20时47分15秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 #include<climits> 25 #include<queue> 26 27 using namespace std; 28 int n , m ; 29 struct node{ 30 int a,b,l,c; 31 }a[100005]; 32 int f[10005]; 33 int hs[10005]; 34 int cmp(struct node a ,struct node b) 35 { 36 if(a.l == b.l) 37 { 38 return a.c < b.c ; 39 } 40 return a.l < b.l; 41 } 42 int find(int x) 43 { 44 return x == f[x]?x:f[x] = find(f[x]); 45 } 46 int main(){ 47 48 while(scanf("%d %d",&n,&m) != EOF){ 49 memset(a,0,sizeof(a)); 50 memset(f,0,sizeof(f)); 51 for(int i = 1;i <= n; i ++) 52 f[i] = i ; 53 for(int i =1 ;i <= m;i ++) 54 { 55 scanf("%d %d %d %d",&a[i].a,&a[i].b,&a[i].l,&a[i].c); 56 } 57 sort(a+1,a+1+m,cmp); 58 int sumc,suml; 59 sumc = suml = 0 ; 60 for(int i = 1;i <= m;i ++) 61 { 62 int ta = find(a[i].a); 63 int tb = find(a[i].b); 64 if(ta != tb) 65 { 66 f[tb] = ta; 67 sumc += a[i].c ; 68 suml += a[i].l ; 69 } 70 } 71 int ok = 0 ; 72 memset(hs,0,sizeof(hs)); 73 for(int i =1;i <= n;i ++) 74 { 75 hs[find(i)] ++ ; 76 if(hs[find(i)] == n) 77 ok = 1; 78 } 79 if(ok) 80 printf("%d %d ",suml,sumc); 81 else printf("-1 -1 "); 82 } 83 return 0 ; 84 }
J:
快捷计算:
1 // File Name: b.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月11日 星期五 19时53分57秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 #include<climits> 25 #include<queue> 26 27 using namespace std; 28 #define mod 1000000007 29 long long num[1200]; 30 long long temp[20][1200]; 31 int main(){ 32 int n; 33 scanf("%d",&n); 34 num[1] = 1; 35 for(int i = 2;i <= n;i ++) 36 { 37 num[i] = num[i-1] * 10 % mod; 38 } 39 for(int i = 0 ;i <= 9 ;i ++) 40 temp[i][1] = i; 41 for(int i = 1;i <= 9;i ++) 42 { 43 for(int j = 2;j <= n;j ++) 44 { 45 temp[i][j] = (temp[i][j-1] + i * num[j])%mod; 46 } 47 } 48 49 long long ans = 0 ; 50 for(int i =0 ;i < n;i ++ ) 51 { 52 for(int j = 0 ;j < n;j ++) 53 { 54 int t; 55 scanf("%1d",&t); 56 ans = (ans + temp[t][n-j]*(j+1) + temp[t][n-i]*(i+1)) % mod; 57 } 58 } 59 printf("%lld ",ans); 60 return 0; 61 }