Others

Uva_125 Because the number of path entries in a-> B is equal to the sum of the product of the number of path entries in a-> I and I-> B, therefore, in the process of using Floyd, we can calculate the number of path lines between A and B. If a-> B

Hdu_1565 There was no idea at the beginning, and later I came to the others' solution report saying that we should construct a bipartite graph and convert it into a minimal cut. Later, I thought about it again. The general principle is as

Uva_104 In fact, this question is to find a minimum ring with the product of exchange rate greater than or equal to 1.01. Because the data volume is small, we can use the dynamic regulation + Floyd to solve the problem. Let's set f [I] [J] [k] to

Hdu_1569 There was no idea at the beginning, and later I came to the others' solution report saying that we should construct a bipartite graph and convert it into a minimal cut. Later, I thought about it again. The general principle is as

Uva_658 This problem has been entangled for a long time. Later, when I made another question, I was told that it is generally faster to use binary to represent the State, in this way, a row can be represented by a number. Continue to think about

Uva_12299 This is the question of Hunan Provincial competition. If I could have learned ytq and gave me a PPT later, it would have been too late. This topic has a shift operation more than the ordinary rmq, store the line segment tree with heap,

Poj_1753 After several days, we have already written this question 3rd Times. After being familiar with bit operations, we have solved this problem more efficiently. At the same time, this question also has a little skill in enumeration. If we

UVA_10057 First, sort the sequence in ascending order. If N is an odd number, A can only be the number in the middle. If N is an even number, then A can be any integer in the interval composed of two numbers. #include#include#includeint

HDU_2066 Directly use the Bellman-Ford algorithm optimized by the queue to find the shortest path of each vertex, and then find the minimum value of the shortest path of each desired location. #include#includeint q[1010],inq[1010];int G[1010][1010],

HDU_2544 Use the Bellman-Ford Algorithm Optimized in the queue to obtain the shortest path of each point. #include#includeint N,M,G[110][110],d[110];int q[110],inq[110];int init(){int i,A,B,C; scanf("%d%d",&N,&M);if(!N&&!M)return 0; memset(G,0

POJ_2723 At the beginning, I still didn't grasp the core of the 2-SAT question and won't make a diagram. In fact, the most important thing about 2-SAT is to find out the relationship between 0 and 1, that is, whether a thing is selected or not, or

HDU_1426 This question seems to be available for deep search, but I actually chose this simple question in the sudoku just to practice the Dancing Links. The basic idea of Dancing Links is to convert the graph into a 0-1 matrix and finally convert

POJ_3683 This is a 2-SAT problem. In fact, algorithms and diagrams are not difficult to think about, but the encoding complexity is relatively high. In the composition, we need to determine whether the two segments are intersecting. the easier way

UVA_12299 First, analyze the simple situation. The number indicates the maximum heap. 2 wins, 3 loses, and 4 wins ,...... In fact, it is clear that even numbers must win, but odd numbers cannot be said, so let's look at it from another angle. If 3

POJ_3468 (1) After digesting the ideas on the PPT, I made a new question. Remember the elements of the original array as a [I], and then we use the heap to create two line segment trees, the original array of a tree is x [I] (x [I] = a [I]-a [I-1],

UVA_10273 Because of the classification of graph theory, I did not dare to write violence at first. Later I found that the problem solution was basically violent, and I wrote it with brute force. The enumerated cycle is the minimum public multiple

Uva_000086 A bare short-circuit problem can be solved directly by using the Bellman-Ford optimized queue. #include#include#define MAXD 20010#define MAXM 100010#define INF 1000000000int N, M, S, T, q[MAXD], inq[MAXD];int first[MAXD], next[MAXM],

UVA_563 We can introduce a source point to be connected to a robbery point, introduce a sink point to be connected to the escaped point, and then check whether the maximum flow is equal to the number of robbery points. Because the capacity of each

HDU_4081 The general idea of this question is like this. First, find a minimum spanning tree, then find or by the way, find the maximum edge weight on the path between any two points on the Minimum Spanning Tree, and then enumerate magic road,

POJ_2182 I did not expect that the line segment tree could be used in this way. It is true that the line segment tree is too flexible ...... This question can be converted into querying from the back to the front. For example, there are x cows whose