CF 1131B Draw!

　　　　　　　　　　　　　　　　　　　　　　　　　Draw!

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

 

Description

You still have partial information about the score during the historic football match. You are given a set of pairs $(ai,bi)(ai,bi),\; indicating\; that\; at\; some\; point\; during\; the\; match\; the\; score\; was\; "aiai: bibi".\; It\; is\; known\; that\; if\; the\; current\; score\; is\; «xx:yy»,\; then\; after\; the\; goal\; it\; will\; change\; to\; "x+1x+1:yy"\; or\; "xx:y+1y+1".\; What\; is\; the\; largest\; number\; of\; times\; a\; draw\; could\; appear\; on\; the\; scoreboard?$

The pairs "${\mathrm{aiai:bibi"\; are\; given\; in\; chronological\; order\; (time\; increases),\; but\; you\; are\; given\; score\; only\; for\; some\; moments\; of\; time.\; The\; last\; pair\; corresponds\; to\; the\; end\; of\; the\; match.}}^{}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 100001\le n\le 10000)\; —\; the\; number\; of\; known\; moments\; in\; the\; match.}$

Each of the next $\mathrm{nn lines\; contains\; integers aiai and bibi (0\le ai,bi\le 1090\le ai,bi\le 109),\; denoting\; the\; score\; of\; the\; match\; at\; that\; moment\; (that\; is,\; the\; number\; of\; goals\; by\; the\; first\; team\; and\; the\; number\; of\; goals\; by\; the\; second\; team).}$

All moments are given in chronological order, that is, sequences ${\mathrm{xixi and yjyj are\; non-decreasing.\; The\; last\; score\; denotes\; the\; final\; result\; of\; the\; match.}}^{}$

Output

Print the maximum number of moments of time, during which the score was a draw. The starting moment of the match (with a score 0:0) is also counted.

Sample Input

Input

3
2 0
3 1
3 4

Output

2

Input

3
0 0
0 0
0 0

Output

1

Input

1
5 4

Output

5

Hint

In the example one of the possible score sequences leading to the maximum number of draws is as follows: 0:0, 1:0, 2:0, 2:1, 3:1, 3:2, 3:3, 3:4.

 

 

 

 題意:

兩個人進行比賽，給出幾場兩人的比分,問在滿足這幾場比分的條件下，最多會出現幾場平局(開始的0:0也算一場）。

 解題思路：

將比賽雙方看成兩條不斷前進的線段，利用區間的方式去模擬比賽的情況。

 重點是Debug，在思路找不到錯誤的情況下，就去多試一些樣例，儘量的做到測試的不重不漏。

 

 

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<string>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<stack>
 9 #include<deque>
10 #include<map>
11 #include<iostream>
12 using namespace std;
13 typedef long long  LL;
14 const double pi=acos(-1.0);
15 const double e=exp(1);
16 const int N = 100009;
17 
18 
19 queue<LL> check;
20 int main()
21 {
22     LL i,p,j,n,t;
23     LL ans = 0;
24     LL a1,b1,a2,b2;
25 
26     while(!check.empty())
27         check.pop();
28 
29     scanf("%lld",&n);
30     scanf("%lld%lld",&a1,&b1);
31 
32     ans += min(a1, b1) + 1;
33     check.push(min(a1,b1));
34     for(i = 1; i < n; i ++)
35     {
36         scanf("%lld%lld",&a2,&b2);
37         if(a2 < b1 || a1 > b2)
38             ;
39         else if(a1 < b1 && a2 <= b2 && a2 >= b1)
40         {
41             ans+=(a2 - b1) + 1;
42 
43 
44             if(check.front() == b1)
45             {
46                 ans--;
47 
48             }
49             check.pop();
50             check.push(a2);
51         }
52         else if(a1 >= b1 && a2 <= b2)
53         {
54             ans+=(a2 - a1 + 1);
55 
56             if(check.front() == a1)
57             {
58                 ans--;
59 
60             }
61             check.pop();
62             check.push(a2);
63         }
64         else if(a1 >= b1 && a1 <= b2 && a2 > b2)
65         {
66             ans += b2 - a1 + 1;
67 
68             if(check.front() == a1)
69             {
70                 ans--;
71 
72             }
73             check.pop();
74             check.push(b2);
75         }
76         else if(a1 <= b1 && a2 >= b2)
77         {
78             ans += b2 - b1 + 1;
79 
80             if(check.front() == b1)
81             {
82                 ans--;
83 
84             }
85             check.pop();
86             check.push(b2);
87         }
88         a1 = a2;
89         b1 = b2;
90     }
91     printf("%lld\n",ans);
92     return 0;
93 }