【HDOJ】1510 White Rectangles

这个题目很好,变形的题目也很多。简单DP。

 1 /* 1510 */
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5
 6 #define MAXN 105
 7
 8 char map[MAXN][MAXN];
 9 int dp[MAXN][MAXN];
10
11 int min(int a, int b) {
12     return a<b ? a:b;
13 }
14
15 int main() {
16     int n;
17     int i, j, k, tmp;
18     int sum;
19
20     #ifndef ONLINE_JUDGE
21         freopen("data.in", "r", stdin);
22     #endif
23
24     while (scanf("%d", &n) != EOF) {
25         for (i=1; i<=n; ++i)
26             scanf("%s", map[i]+1);
27
28         sum = 0;
29         for (i=1; i<=n; ++i) {
30             for (j=1; j<=n; ++j) {
31                 if (map[i][j] == ‘#‘) {
32                     dp[i][j] = 0;
33                 } else {
34                     dp[i][j] = dp[i-1][j] + 1;
35                     tmp = dp[i][j];
36                     sum += tmp;
37                     for (k=j-1; k>0; --k) {
38                         tmp = min(tmp, dp[i][k]);
39                         sum += tmp;
40                     }
41                 }
42             }
43         }
44         printf("%d\n", sum);
45     }
46
47     return 0;
48 }
时间: 2024-11-08 21:26:22

【HDOJ】1510 White Rectangles的相关文章

【HDOJ】4956 Poor Hanamichi

基本数学题一道,看错位数,当成大数减做了,而且还把方向看反了.所求为最接近l的值. 1 #include <cstdio> 2 3 int f(__int64 x) { 4 int i, sum; 5 6 i = sum = 0; 7 while (x) { 8 if (i & 1) 9 sum -= x%10; 10 else 11 sum += x%10; 12 ++i; 13 x/=10; 14 } 15 return sum; 16 } 17 18 int main() { 1

【HDOJ】1099 Lottery

题意超难懂,实则一道概率论的题目.求P(n).P(n) = n*(1+1/2+1/3+1/4+...+1/n).结果如果可以除尽则表示为整数,否则表示为假分数. 1 #include <cstdio> 2 #include <cstring> 3 4 #define MAXN 25 5 6 __int64 buf[MAXN]; 7 8 __int64 gcd(__int64 a, __int64 b) { 9 if (b == 0) return a; 10 else return

【HDOJ】2844 Coins

完全背包. 1 #include <stdio.h> 2 #include <string.h> 3 4 int a[105], c[105]; 5 int n, m; 6 int dp[100005]; 7 8 int mymax(int a, int b) { 9 return a>b ? a:b; 10 } 11 12 void CompletePack(int c) { 13 int i; 14 15 for (i=c; i<=m; ++i) 16 dp[i]

【HDOJ】3509 Buge&#39;s Fibonacci Number Problem

快速矩阵幂,系数矩阵由多个二项分布组成.第1列是(0,(a+b)^k)第2列是(0,(a+b)^(k-1),0)第3列是(0,(a+b)^(k-2),0,0)以此类推. 1 /* 3509 */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #incl

【HDOJ】1818 It&#39;s not a Bug, It&#39;s a Feature!

状态压缩+优先级bfs. 1 /* 1818 */ 2 #include <iostream> 3 #include <queue> 4 #include <cstdio> 5 #include <cstring> 6 #include <cstdlib> 7 #include <algorithm> 8 using namespace std; 9 10 #define MAXM 105 11 12 typedef struct {

【HDOJ】2424 Gary&#39;s Calculator

大数乘法加法,直接java A了. 1 import java.util.Scanner; 2 import java.math.BigInteger; 3 4 public class Main { 5 public static void main(String[] args) { 6 Scanner cin = new Scanner(System.in); 7 int n; 8 int i, j, k, tmp; 9 int top; 10 boolean flag; 11 int t

【HDOJ】2425 Hiking Trip

优先级队列+BFS. 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 using namespace std; 6 7 #define MAXN 25 8 9 typedef struct node_st { 10 int x, y, t; 11 node_st() {} 12 node_st(int xx, int yy, int tt)

【HDOJ】1686 Oulipo

kmp算法. 1 #include <cstdio> 2 #include <cstring> 3 4 char src[10005], des[1000005]; 5 int next[10005], total; 6 7 void kmp(char des[], char src[]){ 8 int ld = strlen(des); 9 int ls = strlen(src); 10 int i, j; 11 12 total = i = j = 0; 13 while (

【HDOJ】2795 Billboard

线段树.注意h范围(小于等于n). 1 #include <stdio.h> 2 #include <string.h> 3 4 #define MAXN 200005 5 #define lson l, mid, rt<<1 6 #define rson mid+1, r, rt<<1|1 7 #define mymax(x, y) (x>y) ? x:y 8 9 int nums[MAXN<<2]; 10 int h, w; 11 12