poj-1083 贪心法版本

来源：互联网发布：冯.霍恩海姆知乎编辑：程序博客网时间：2024/04/25 10:22

#include <stdio.h>
#include <stdlib.h>

struct tableMove{
   int src;
   int dst;
};

int getMax(int a, int b) {
   return a >= b ? a : b;
}

int getMin(int a, int b) {
   return a < b ? a : b;
}

void tableMoveswap(tableMove * moveA, tableMove * moveB) {
   int tmpSrc = moveA->src;
   int tmpDst = moveA->dst;
   moveA->src = moveB->src;
   moveA->dst = moveB->dst;
   moveB->src = tmpSrc;
   moveB->dst = tmpDst;
}

void sortTableMoveArrayAsDst(tableMove * tabelMoveArray, int length) {
   for (int i = 0; i < length - 1; i++) {
       bool exchanged = false;
       for (int j = 0; j <= (length - i -1) - 1; j++) {
           if (tabelMoveArray[j].dst > tabelMoveArray[j+1].dst) {
               tableMoveswap(&tabelMoveArray[j], &tabelMoveArray[j+1]);
               exchanged = true;
           }
       }
       if (exchanged == false) {
           break;
       }
   }
}

void sortTableMoveArrayAsSrc(tableMove * tabelMoveArray, int length) {
   for (int i = 0; i < length - 1; i++) {
       bool exchanged = false;
       for (int j = 0; j <= (length - i -1) - 1; j++) {
           if (tabelMoveArray[j].src > tabelMoveArray[j+1].src) {
               tableMoveswap(&tabelMoveArray[j], &tabelMoveArray[j+1]);
               exchanged = true;
           }
       }
       if (exchanged == false) {
           break;
       }
   }
}

int getMinMoveTime(tableMove * tabelMoveArray, int tabelMoveTime) {
   sortTableMoveArrayAsSrc(tabelMoveArray, tabelMoveTime);
   int moveTime = 0;
   int SUM = tabelMoveTime;
   while(1) {
       int dst = -1;
       for (int i = 0; i < tabelMoveTime; i++) {
           if (tabelMoveArray[i].dst > 0 && dst == -1) {
               dst = tabelMoveArray[i].dst;
               tabelMoveArray[i].dst = -1;
               SUM--;
           } else if (tabelMoveArray[i].dst > 0) {
               if (tabelMoveArray[i].src > dst) {
                   dst = tabelMoveArray[i].dst;
                   tabelMoveArray[i].dst = -1;
                   SUM--;
               }
           }
       }
       moveTime++;
       if (SUM == 0) {
           break;
       }
   }
   return moveTime;
}

void runOneCase() {
   int tabelMoveTime = 0;
   scanf("%d", &tabelMoveTime);
   tableMove * tabelMoveArray= (tableMove *)malloc(sizeof(tableMove) * tabelMoveTime);
   for (int i = 0; i < tabelMoveTime; i++) {
       int src, dst = 0;
       scanf("%d %d", &src, &dst);
       if (src % 2 == 1) {
           src++;
       }
       if (dst % 2 == 1) {
           dst++;
       }
       tabelMoveArray[i].src = getMin(src, dst);
       tabelMoveArray[i].dst = getMax(src, dst);
   }
   int minMoveTime = getMinMoveTime(tabelMoveArray, tabelMoveTime);
   printf("%d\n", minMoveTime * 10);
   free(tabelMoveArray);
   tabelMoveArray = NULL;
}

int main() {
   int testCaseNum = 0;
   scanf("%d", &testCaseNum);
   for (int i = 0; i < testCaseNum; i++) {
       runOneCase();
   }

}

用贪心法的版本，

看了discuss里的另一种解答才明白为何此题能算水题（回头写一下，也要记住）。

疑问:

1. 按照 dst 进行排序得到的一直不对，按照 src 就对了，要好好看看贪心法了。

与算法导论中的贪心法实例的不同在于, 该题要求的是一个全局的最优（全部移动完所需的总次数），而实例一般求的是一次移动能移动最多的个数。

有很大的不同，上面按照src排序，再用传统贪心法，能得到答案，但是如何证明？

0 0