uva 11111 Generalized Matrioshkas

这道题题意有点难懂，其实问题可以转换为求区间是否相交，以及包含关系是否正确，没什么算法的东西，写程序的时候小心注意不要出现bug就行了，注意判断输入数据错误的条件还有配对不成功的情况，题目不难，就是要把代码调好有点消耗时间。

#include <stdio.h>#include <vector>#include <map>#include <algorithm>using namespace std;#defineMAX_SIZE3000typedef int START_POS;typedef int INDEX;map<START_POS,INDEX> m;int v[MAX_SIZE*2];struct node{int value;int start;int end;};struct node arr[MAX_SIZE];int cmp(const void *a, const void *b){struct node *pa = (struct node*)a;struct node *pb = (struct node*)b;return pa->start - pb->start;}void func(int len){int i, j;int count;bool f1, f2, f3;int start_pos;int sum_value;if(len%2){printf(":-( Try again.\n");return;}count = 0;for(i=1; i<len; i++){if(v[i] < 0){for(j=i+1; j<=len; j++)if(v[i] == -1*v[j]){arr[count].start = i;arr[count].end = j;arr[count].value = v[j];count ++;v[j] = 0;break;}}}if(count != len/2){printf(":-( Try again.\n");return;}qsort((void*)arr, count, sizeof(struct node), cmp);m.clear();for(i=0; i<count; i++)m[arr[i].start] = i;//开始检查是否正确for(i=0; i<count; i++){f1 = f2 = f3 = true;sum_value = 0;for(j=i+1; j<count; ){if(arr[j].start > arr[i].end)break;//检查区间有没有交叉的情况出现if(arr[j].end > arr[i].end){f1 = false;goto judge;}//检查包含的区间的权值是不是小于arr[i]的权值if(arr[j].value >= arr[i].value){f2 = false;goto judge;}sum_value +=  arr[j].value;start_pos = arr[j].end+1;while(1){if(start_pos >= arr[i].end){j = count;break;}if(m.find(start_pos) != m.end()){j = m[start_pos];break;}start_pos ++;}}if(sum_value >= arr[i].value){f3 = false;goto judge;}judge:if(!f1 || !f2 || !f3){printf(":-( Try again.\n");return;}}printf(":-) Matrioshka!\n");}int main(void){int x, count;int res;char ch;while(1){count = 1;while(1){res = scanf("%d", &x);if(res == EOF)goto end;v[count++] = x;ch = getchar();if(ch == '\n'){func(count-1);break;}}}end:return 0;}