【POJ1733】【带标记并查集】Parity game

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <vector>
  6 #include <utility>
  7 #include <iomanip>
  8 #include <string>
  9 #include <cmath>
 10 #include <queue>
 11 #include <map>
 12 
 13 const int MAXN = 5000 + 10; 
 14 const int MAX = 32000 + 10; 
 15 using namespace std;
 16 struct DATA{
 17        int num, order;
 18        bool operator < (DATA b)const{
 19             return num < b.num;
 20        }
 21 }data[MAXN * 2];//data记录的是出现过的数字
 22 struct DATA2{
 23        int a, b, type;//表示sum[b] - sum[a - 1]的是否是同一类型 
 24 }data2[MAXN];//直接的输入数据 
 25 int parent[MAXN * 2];
 26 int val[MAXN * 2];
 27 
 28 int n, q, cnt;
 29 
 30 void init(){
 31      scanf("%d%d", &n, &q);
 32      for (int i = 1; i <= q; i++){
 33           char str[10];
 34           scanf("%d%d", &data2[i].a, &data2[i].b);
 35           data[i * 2 - 1].num = data2[i].a; data[i * 2 - 1].order = i * 2 - 1;
 36           data[i * 2].num = data2[i].b; data[i * 2].order = i * 2;
 37           scanf("%s", str);
 38           //odd是奇数，even是偶数 
 39           if (str[0] == ‘o‘) data2[i].type = 0;
 40           else if (str[0] == ‘e‘) data2[i].type = 1; 
 41      }
 42      //准备离散化
 43      sort(data + 1, data + 1 + 2 * q);
 44      cnt = 0;//记录数字的 
 45      data[0].num = -100;
 46      for (int i = 1; i <= 2 * q; i++){
 47          if (data[i].num != data[i - 1].num) ++cnt;
 48          if (data[i].order % 2 == 1) data2[((data[i].order - 1) / 2) + 1].a = cnt;
 49          else data2[((data[i].order - 1) / 2) + 1].b = cnt;
 50      }  
 51      /*for (int i = 1;i <= q; i++){
 52          printf("%d %d %d\n", data2[i].a, data2[i].b, data2[i].type);
 53      }*/
 54 }
 55 int find(int x){
 56     int tmp = 0, f = x;
 57     //注意要加上路径压缩 
 58     while (x != parent[x]){
 59           tmp += val[x];
 60           x = parent[x];
 61     }
 62     parent[f] = x;
 63     val[f] = tmp % 2;
 64     return x;  
 65 }
 66 
 67 void work(){
 68      //val[i] = 1则与父亲的奇偶性不同，0则相同 
 69      for (int i = 1; i <= cnt; i++){
 70          parent[i] = i;
 71          val[i] = 0;
 72      }
 73      for (int i = 1; i <= q; i++){
 74          int x = data2[i].a, y = data2[i].b ,t = data2[i].type;
 75          y++;
 76          if (i == 2)
 77          printf("");
 78          int xx = find(x), yy = find(y), flag = 0;
 79          if (xx == yy){//同根 
 80             if (t == 1){//需要相同 
 81                if (val[x] != val[y]) flag = 1;
 82                //相同则躲过一劫 
 83             }else{//需要不同 
 84                if (val[x] == val[y]) flag = 1;
 85             }
 86          }else{//不管怎么样都要合并 
 87             if (t == 1){
 88                parent[xx] = y;
 89                val[xx] = (-val[x] + 2) % 2;
 90             }else{
 91                parent[xx] = y;
 92                val[xx] = (-val[x] + 3) % 2;
 93             }
 94          }
 95          if (flag == 1) {printf("%d\n", i - 1);return;}
 96      }
 97      printf("%d\n", q);
 98 }
 99 
100 int main(){
101     int T;
102     #ifdef LOCAL
103     freopen("data.txt",  "r",  stdin);
104     freopen("out.txt",  "w",  stdout); 
105     #endif
106     init();
107     work();
108     return 0;
109 }