Poj--1009（模拟，技巧）
2014-11-27 02:44:34
思路：这题写了好久....感觉最后也没写的好。
处理方法就是处理每段的起点，及其周围8个点的值，最后将答案按位置排序，中间的位置的值和离他最近的前面的答案值一样，注意还要处理总长+1的位置！（思考）
二分这个问题，还是要考虑清楚，比较特殊的二分要在while里面加特判条件。
1 /************************************************************************* 2 > File Name: p1009.cpp 3 > Author: Nature 4 > Mail: 564374850@qq.com 5 > Created Time: Tue 25 Nov 2014 07:41:53 PM CST 6 ************************************************************************/ 7 8 #include <cstdio> 9 #include <cstring> 10 #include <cstdlib> 11 #include <cmath> 12 #include <vector> 13 #include <map> 14 #include <set> 15 #include <stack> 16 #include <queue> 17 #include <iostream> 18 #include <algorithm> 19 using namespace std; 20 #define lp (p << 1) 21 #define rp (p << 1|1) 22 #define getmid(l,r) (l + (r - l) / 2) 23 #define MP(a,b) make_pair(a,b) 24 typedef long long ll; 25 const int INF = 1 << 30; 26 27 struct node{ 28 int pos,val; 29 node(int a,int b) : pos(a),val(b) {} 30 }; 31 32 vector<node> mp; 33 vector<node> ans; 34 int n,v[1010],len[1010],sum; 35 36 int B_search(int val){ 37 int l = 0,r = mp.size() - 1; 38 while(l <= r){ 39 int mid = getmid(l,r); 40 if(l + 1 >= r){ 41 if(mp[r].pos <= val) return mp[r].val; 42 else return mp[l].val; 43 } 44 if(mp[mid].pos <= val) 45 l = mid; 46 else 47 r = mid - 1; 48 } 49 return mp[l].val; 50 } 51 52 void Cal(int num,int val){ 53 if(num > sum){ 54 return; 55 } 56 int res = 0; 57 if(num > n){ 58 res = max(res,abs(val - B_search(num - n))); 59 if((num - n) % n != 0) res = max(res,abs(val - B_search(num - n + 1))); 60 if((num - n) % n != 1) res = max(res,abs(val - B_search(num - n - 1))); 61 } 62 if(num % n != 0) res = max(res,abs(val - B_search(num + 1))); 63 if(num % n != 1) res = max(res,abs(val - B_search(num - 1))); 64 if(num + n <= sum){ 65 res = max(res,abs(val - B_search(num + n))); 66 if((num + n) % n != 0) res = max(res,abs(val - B_search(num + n + 1))); 67 if((num + n) % n != 1) res = max(res,abs(val - B_search(num + n - 1))); 68 } 69 //printf("Plus : %d %d\n",num,res); 70 ans.push_back(node(num,res)); //position and maximum absolute value. 71 } 72 73 void Solve(int num,int val){ 74 int tmp; 75 Cal(num,val); 76 if(num > n){ 77 tmp = B_search(num - n); 78 Cal(num - n,tmp); 79 if((num - n) % n != 0){ 80 tmp = B_search(num - n + 1); 81 Cal(num - n + 1,tmp); 82 } 83 if((num - n) % n != 1){ 84 tmp = B_search(num - n - 1); 85 Cal(num - n - 1,tmp); 86 } 87 } 88 if(num % n != 0){ 89 tmp = B_search(num + 1); 90 Cal(num + 1,tmp); 91 } 92 if(num % n != 1){ 93 tmp = B_search(num - 1); 94 Cal(num - 1,tmp); 95 } 96 if(num + n <= sum){ 97 tmp = B_search(num + n); 98 Cal(num + n,tmp); 99 if((num + n) % n != 0){ 100 tmp = B_search(num + n + 1); 101 Cal(num + n + 1,tmp); 102 } 103 if((num + n) % n != 1){ 104 tmp = B_search(num + n - 1); 105 Cal(num + n - 1,tmp); 106 } 107 } 108 } 109 110 bool cmp(node a,node b){ 111 return a.pos < b.pos; 112 } 113 114 int main(){ 115 int a,b; 116 while(scanf("%d",&n) != EOF && n){ 117 mp.clear(); 118 ans.clear(); 119 sum = 0; 120 int pos = 1,cnt = 0; 121 while(scanf("%d%d",&a,&b) != EOF){ 122 if(!a && !b) 123 break; 124 sum += b; 125 mp.push_back(node(pos,a)); 126 mp.push_back(node(pos + b - 1,a)); 127 pos += b; 128 v[++cnt] = a; 129 len[cnt] = b; 130 } 131 int tot = 1; 132 for(int i = 1; i <= cnt; ++i){ 133 Solve(tot,v[i]); 134 tot += len[i]; 135 } 136 Solve(tot,0); 137 sort(ans.begin(),ans.end(),cmp); 138 int pre = 1,sz = ans.size(); 139 printf("%d\n",n); 140 for(int i = 0; i < sz - 1; ++i){ 141 if(ans[i].val != ans[i + 1].val){ 142 printf("%d %d\n",ans[i].val,ans[i + 1].pos - pre); 143 pre = ans[i + 1].pos; 144 } 145 } 146 printf("%d %d\n",ans[sz - 1].val,sum - pre + 1); 147 printf("0 0 \n"); 148 } 149 printf("0\n"); 150 return 0; 151 }
