【HDU】4391 Paint The Wall

题意：

给出n个节点，编号0~n-1。

1 把[x,y]颜色覆盖为z。

2 询问[x,y]颜色为z的有多少个。

 

题解：http://page.renren.com/601081183/note/867254911

懒得hash，直接用map搞。

把数列分成sqrt(n)段，如果每段在完整的查询区间内，可以O(1)得到答案，否则对该段暴力答案。

  1 #include<cstdio>
  2 #include<cmath>
  3 #include<map>
  4 #define EPS 1e-8
  5 #define MAXM 331
  6 #define MAXN 100010
  7 using namespace std;
  8 int col[MAXN], same[MAXM];
  9 map<int, int> mymap[MAXM];
 10 int n, block;
 11 void Hash(int t) {
 12     int nd, i;
 13     mymap[t].clear();
 14     nd = min(t * block + block, n);
 15     for (i = t * block; i < nd; i++)
 16         mymap[t][col[i]]++;
 17 }
 18 void PushDown(int t) {
 19     int i, nd;
 20     nd = min(t * block + block, n);
 21     if (same[t] >= 0) {
 22         for (i = t * block; i < nd; i++)
 23             col[i] = same[t];
 24         same[t] = -1;
 25     }
 26 }
 27 void Update(int x, int y, int z) {
 28     int t, i, nd;
 29     t = x / block;
 30     if (t == y / block) {
 31         if (same[t] != z) {
 32             PushDown(t);
 33             for (i = x; i <= y; i++)
 34                 col[i] = z;
 35             Hash(t);
 36         }
 37     } else {
 38         if (x % block) {
 39             if (same[t] != z) {
 40                 PushDown(t);
 41                 for (nd = min(t * block + block, n); x < nd; x++)
 42                     col[x] = z;
 43                 Hash(t);
 44             } else
 45                 x = min(t * block + block, n);
 46         }
 47         if (y % block != block - 1) {
 48             t = y / block;
 49             if (same[t] != z) {
 50                 PushDown(t);
 51                 for (nd = t * block; y >= nd; y--)
 52                     col[y] = z;
 53                 Hash(t);
 54             } else
 55                 y = t * block - 1;
 56         }
 57         for (t = x / block; x <= y; x += block, t++)
 58             same[t] = z;
 59     }
 60 }
 61 int Query(int x, int y, int z) {
 62     int t, i, nd, ans;
 63     t = x / block;
 64     ans = 0;
 65     if (t == y / block) {
 66         if (same[t] == z)
 67             ans += y - x + 1;
 68         else if (same[t] == -1) {
 69             for (i = x; i <= y; i++) {
 70                 if (col[i] == z)
 71                     ans++;
 72             }
 73         }
 74     } else {
 75         if (x % block) {
 76             if (same[t] == z) {
 77                 ans += block - x % block;
 78                 x = min(t * block + block, n);
 79             } else if (same[t] == -1) {
 80                 for (nd = min(t * block + block, n); x < nd; x++) {
 81                     if (col[x] == z)
 82                         ans++;
 83                 }
 84             } else
 85                 x = min(t * block + block, n);
 86         }
 87         if (y % block != block - 1) {
 88             t = y / block;
 89             if (same[t] == z) {
 90                 ans += y % block + 1;
 91                 y = t * block - 1;
 92             } else if (same[t] == -1) {
 93                 for (nd = t * block; y >= nd; y--) {
 94                     if (col[y] == z)
 95                         ans++;
 96                 }
 97             } else
 98                 y = t * block - 1;
 99         }
100         for (t = x / block; x <= y; x += block, t++) {
101             if (same[t] >= 0) {
102                 if (same[t] == z)
103                     ans += block;
104             } else {
105                 if (mymap[t].count(z))
106                     ans += mymap[t][z];
107             }
108         }
109     }
110     return ans;
111 }
112 int main() {
113     int q, i, j, cmd, x, y, z;
114     while (~scanf("%d%d", &n, &q)) {
115         block = (int) (ceil(sqrt((double) n)) + EPS);
116         for (i = 0; i < n; i++)
117             scanf("%d", &col[i]);
118         for (i = j = 0; i < n; i += block, j++) {
119             same[j] = -1;
120             Hash(j);
121         }
122         while (q--) {
123             scanf("%d%d%d%d", &cmd, &x, &y, &z);
124             if (cmd == 1)
125                 Update(x, y, z);
126             else
127                 printf("%d\n", Query(x, y, z));
128         }
129     }
130     return 0;
131 }
posted on 2012-08-27 17:08  DrunBee  阅读(337)  评论(0)  编辑  收藏  举报