PKU ACM 1080 源代码
2011-01-03 00:25
274 查看
1080是LCS问题,关键是CreatLCSMatrix()的设计。
/*Human Gene Functions*/ /*Description It is well known that a human gene can be considered as a sequence, consisting of four nucleotides, which are simply denoted by four letters, A, C, G, and T. Biologists have been interested in identifying human genes and determining their functions, because these can be used to diagnose human diseases and to design new drugs for them. A human gene can be identified through a series of time-consuming biological experiments, often with the help of computer programs. Once a sequence of a gene is obtained, the next job is to determine its function. One of the methods for biologists to use in determining the function of a new gene sequence that they have just identified is to search a database with the new gene as a query. The database to be searched stores many gene sequences and their functions – many researchers have been submitting their genes and functions to the database and the database is freely accessible through the Internet. A database search will return a list of gene sequences from the database that are similar to the query gene. Biologists assume that sequence similarity often implies functional similarity. So, the function of the new gene might be one of the functions that the genes from the list have. To exactly determine which one is the right one another series of biological experiments will be needed. Your job is to make a program that compares two genes and determines their similarity as explained below. Your program may be used as a part of the database search if you can provide an efficient one. Given two genes AGTGATG and GTTAG, how similar are they? One of the methods to measure the similarity of two genes is called alignment. In an alignment, spaces are inserted, if necessary, in appropriate positions of the genes to make them equally long and score the resulting genes according to a scoring matrix. For example, one space is inserted into AGTGATG to result in AGTGAT-G, and three spaces are inserted into GTTAG to result in –GT--TAG. A space is denoted by a minus sign (-). The two genes are now of equal length. These two strings are aligned: AGTGAT-G -GT--TAG In this alignment, there are four matches, namely, G in the second position, T in the third, T in the sixth, and G in the eighth. Each pair of aligned characters is assigned a score according to the following scoring matrix. denotes that a space-space match is not allowed. The score of the alignment above is (-3)+5+5+(-2)+(-3)+5+(-3)+5=9. Of course, many other alignments are possible. One is shown below (a different number of spaces are inserted into different positions): AGTGATG -GTTA-G This alignment gives a score of (-3)+5+5+(-2)+5+(-1) +5=14. So, this one is better than the previous one. As a matter of fact, this one is optimal since no other alignment can have a higher score. So, it is said that the similarity of the two genes is 14. Input The input consists of T test cases. The number of test cases ) (T is given in the first line of the input file. Each test case consists of two lines: each line contains an integer, the length of a gene, followed by a gene sequence. The length of each gene sequence is at least one and does not exceed 100. Output The output should print the similarity of each test case, one per line. Sample Input 2 7 AGTGATG 5 GTTAG 7 AGCTATT 9 AGCTTTAAA Sample Output 14 21 */ #include "stdlib.h" #include "stdio.h" #include "string.h" #define LEFT 1 #define UP 2 #define LEFT_UP 3 #define MAX_STRING_LEN 255 int aiScoringMatrix[5][5] = {{5,-1,-2,-1,-3},/ {-1,5,-3,-2,-4},/ {-2,-3,5,-2,-2},/ {-1,-2,-2,5,-1},/ {-3,-4,-2,-1,0}}; char gacB[MAX_STRING_LEN][MAX_STRING_LEN] = {0}; int gaiComSubseqLen[MAX_STRING_LEN][MAX_STRING_LEN] = {0}; #define GETINDEX(vcInputChar,iScoreIndex) { if ('A' == vcInputChar ||'a' == vcInputChar ) { iScoreIndex = 0; }/ else if ('C' == vcInputChar ||'c' == vcInputChar ) {iScoreIndex = 1; }/ else if ('G' == vcInputChar ||'g' == vcInputChar ) {iScoreIndex = 2; }/ else/*T*/ {iScoreIndex = 3; }} #define GETMAXSCORE(iScoreUp,iScoreLeft,iScoreLeftUp) {iMaxScore = iScoreLeftUp > iScoreUp?iScoreLeftUp:iScoreUp;/ iMaxScore = iMaxScore > iScoreLeft?iMaxScore:iScoreLeft;} #define NEGATIVEINFINITY (-9999) void PrintLCSMatrix(char *vpcStringX,char *vpcStringY) { int iLoopX = 0; int iLoopY = 0; int iScoreUp,iScoreLeft,iScoreLeftUp,iMaxScore; int iScoreXIndex = 0; int iScoreYIndex = 0; int iStringXLen = strlen(vpcStringX); int iStringYLen = strlen(vpcStringY); for (iLoopX = 0; iLoopX < iStringXLen; iLoopX++) { for (iLoopY = 0; iLoopY < iStringYLen; iLoopY++) { printf("%3d ",gaiComSubseqLen[iLoopX+1][iLoopY+1]); } printf("/n"); } return; } void CreatLCSMatrix(char *vpcStringX,char *vpcStringY) { int iLoop = 0; int iLoopX = 0; int iLoopY = 0; int iScoreUp,iScoreLeft,iScoreLeftUp,iMaxScore; int iScoreXIndex = 0; int iScoreYIndex = 0; int iStringXLen = strlen(vpcStringX); int iStringYLen = strlen(vpcStringY); for (iLoopX = 0; iLoopX < iStringXLen; iLoopX++) { for (iLoopY = 0; iLoopY < iStringYLen; iLoopY++) { GETINDEX(vpcStringX[iLoopX],iScoreXIndex); GETINDEX(vpcStringY[iLoopY],iScoreYIndex); if (0 == iLoopX && 0 == iLoopY) { iScoreUp = aiScoringMatrix[iScoreXIndex][4]; iScoreLeft = aiScoringMatrix[4][iScoreYIndex]; iScoreLeftUp = aiScoringMatrix[iScoreXIndex][iScoreYIndex]; } else if (0 == iLoopX ) { iScoreUp = NEGATIVEINFINITY; iScoreLeft = aiScoringMatrix[4][iScoreYIndex]; iScoreLeftUp = aiScoringMatrix[iScoreXIndex][iScoreYIndex]; for (iLoop = 0; iLoop < iLoopY; iLoop++) { GETINDEX(vpcStringY[iLoopY-iLoop-1],iScoreYIndex); iScoreLeft += aiScoringMatrix[4][iScoreYIndex]; iScoreLeftUp += aiScoringMatrix[4][iScoreYIndex]; } } else if (0 == iLoopY) { iScoreUp = aiScoringMatrix[iScoreXIndex][4]; iScoreLeft = NEGATIVEINFINITY; iScoreLeftUp = aiScoringMatrix[iScoreXIndex][iScoreYIndex]; for (iLoop = 0; iLoop < iLoopX; iLoop++) { GETINDEX(vpcStringX[iLoopX-iLoop-1],iScoreXIndex); iScoreUp += aiScoringMatrix[iScoreXIndex][4]; iScoreLeftUp += aiScoringMatrix[iScoreXIndex][4]; } } else { iScoreUp = aiScoringMatrix[iScoreXIndex][4] + gaiComSubseqLen[iLoopX][iLoopY+1]; iScoreLeft = aiScoringMatrix[4][iScoreYIndex]+ gaiComSubseqLen[iLoopX+1][iLoopY]; iScoreLeftUp = aiScoringMatrix[iScoreXIndex][iScoreYIndex]+ gaiComSubseqLen[iLoopX][iLoopY]; } GETMAXSCORE(iScoreUp,iScoreLeft,iScoreLeftUp); gaiComSubseqLen[iLoopX+1][iLoopY+1] = iMaxScore; } } return; } void main() { int iNumberOfTestCases; int iLoopCase; int iScore= 0; char acStringX[MAX_STRING_LEN]; char acStringY[MAX_STRING_LEN]; int iStringXLen = 0; int iStringYLen = 0; scanf("%d",&iNumberOfTestCases); for(iLoopCase = 0; iLoopCase < iNumberOfTestCases; iLoopCase++) { scanf("%d %s",&iStringXLen,acStringX); scanf("%d %s",&iStringYLen,acStringY); if (iStringXLen >= MAX_STRING_LEN-1 ||iStringYLen >= MAX_STRING_LEN-1) { printf("string is too long!/n"); return; } memset(gacB,0,MAX_STRING_LEN*MAX_STRING_LEN*sizeof(char)); memset(gaiComSubseqLen,0,MAX_STRING_LEN*MAX_STRING_LEN*sizeof(int)); CreatLCSMatrix(acStringX,acStringY); //PrintLCSMatrix(acStringX,acStringY); iScore = gaiComSubseqLen[iStringXLen][iStringYLen] ; printf("%d/n",iScore); } return; }
相关文章推荐
- Pku acm 1080 Humman Gene Function 动态规划题目解题报告(八)
- PKU ACM 题目1018 源代码
- PKU ACM 1164 源代码
- PKU ACM poj 2231 源代码
- PKU ACM 1080-human gene function
- Pku acm 1014 Dividing 动态规划题目解题报告(十七)
- Pku acm 1125 Stockbroker Grapevine 数据结构题目解题报告(八)---- 弗洛伊德(floyd)算法
- ACM pku 3414 Pots 关于bfs的一个好题
- acm pku 1218 The drunk jailer的算法分析与实现
- pku ACM 1125
- Pku acm 1961 Period数据结构题目解题报告(十九)----kmp算法
- POJ 1006Biorhythms解题报告——生理周期——【PKU ACM】
- ACM基本算法分类、推荐学习资料和配套pku习题
- http://acm.pku.edu.cn/JudgeOnline/
- PKUACM_HUMAN GENE FUNCTION
- Flip and Shift -- ACM PKU 1063 解题报告
- Pku acm 2159 Ancient Cipher 排序算法解题报告(六)----计数排序
- pku-acm
- http://acm.pku.edu.cn/JudgeOnline/problem?id=2480 欧拉
- Pku acm 2406 Power Strings数据结构题目解题报告(十八)----kmp算法