【MPI编程】任意节点数的蝶形求和（高性能计算）

川长思鸟来 2022-04-18 01:54 178阅读 0赞

## 简述 ##

对于任意情况下的蝶形求和，本质上和任意的树形求和是一样的。只需要做逻辑上的抽象，假设那些点存在补全就好了，然后再加一个限定条件就ok了

## 代码 ##

#include<stdio.h>
    #include<string.h>
    #include<mpi.h>
    #pragma warning(disable : 4996)
    #define MAX_STRING 100
    using namespace std;
    #include <fstream>
    #include <iostream>
    
    int main(void) { 
    	int len;
    	double scalar, local_sum = 0;
    	int comm_sz;
    	int my_rank;
    	int divided_len;
    	int last_divided;
    	int begin_i;
    	int oper_N = 2;
    	MPI_Init(NULL, NULL);
    	MPI_Comm_size(MPI_COMM_WORLD, &comm_sz);
    	MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
    	
    	
    	// 只有一个线程的时候不操作
    	if (comm_sz <= 1) { 
    		MPI_Finalize();
    		return 0;
    	}
    	ifstream cin("D:\\C++\\VS\\repo\\MPI-DEMO\\MPI-DEMO\\input.txt");
    	cin >> len; // 输入数据长度
    
    	divided_len = len / comm_sz;
    	last_divided = len % comm_sz;
    
    	if (my_rank < last_divided) { 
    		divided_len++;
    		begin_i = my_rank * divided_len;
    	}
    	else { 
    		begin_i = (my_rank - last_divided) * divided_len + ((divided_len + 1)* last_divided);
    	}
    	// 局部和
    	for (int i = 0; i < len; ++i) { 
    		cin >> scalar;
    		if (i < begin_i || i >= begin_i + divided_len) continue;
    		else { 
    			local_sum += scalar;
    		}
    	}
    	// 假设comm_sz不一定是2的幂
    	// 蝶形求和
    	while (oper_N <= comm_sz || (oper_N > comm_sz && oper_N / 2 < comm_sz)) { 
    		// 往前发
    		if (my_rank % oper_N < oper_N / 2) { 
    			if (my_rank + oper_N / 2 < comm_sz){ 
    				MPI_Send(&local_sum, 1, MPI_DOUBLE, my_rank + oper_N / 2, 0, MPI_COMM_WORLD);
    				MPI_Recv(&scalar, 1, MPI_DOUBLE, my_rank + oper_N / 2, 0, MPI_COMM_WORLD, MPI_STATUSES_IGNORE);
    				local_sum += scalar;
    			}
    		}// 往后发
    		else if (my_rank % oper_N >= oper_N / 2) { 
    			MPI_Send(&local_sum, 1, MPI_DOUBLE, my_rank - oper_N / 2, 0, MPI_COMM_WORLD);
    			MPI_Recv(&scalar, 1, MPI_DOUBLE, my_rank - oper_N / 2, 0, MPI_COMM_WORLD, MPI_STATUSES_IGNORE);
    			local_sum += scalar;
    		}
    		oper_N *= 2;
    	}
    	if (my_rank == 0) { 
    		cout << local_sum << endl;
    	}
    	MPI_Finalize();
    
    	return 0;
    }