C

C - sui题/经典问题
You are given two interval collections A and B. Collection A has N intervals [ A1, A2 ], [ A3, A4 ], …, [ A2N-1, A2N ] and collection B has M intervals [ B1, B2 ], [ B3, B4 ], …, [ B2M-1, B2M ]. Your task is to calculate the length of A - B.

For example if A = {[2, 5], [4, 10], [14, 18]} and B = {[1, 3], [8, 15]}, the length of A - B ({(3, 8), (15, 18]}) is 8.

Input
Line 1: Two integers N and M (1 ≤ N, M ≤ 100000).

Line 2: 2*N integers, A1, A2, …, A2N (1 ≤ Ai ≤ 100000000).

Line 3: 2*M integers, B1, B2, …, B2M (1 ≤= Bi ≤ 100000000).

Output
The length of A - B.

Sample Input
3 2
2 5 4 10 14 18
1 3 8 15
Sample Output
8
我又写了一份毒代码，能够感觉感觉思路是对，但就是wa，调试了好久才发现最后区间减的时候有两种情况没考虑进去，哎，水平没达到，希望自己和别人少犯这种错误。

思路：先对区间排序，之后对区间合并，最后相减。

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int N = 100050;int n,m;struct node{    int a, b;}s1[N],s2[N];bool cmp(node a,node b){    return a.a<b.a||a.a==b.a&&a.b<b.b;}int flow(node *p,int h,int g){    int a=1,c=g;    for(int i=2;i<=g;i++)    {        if(p[a].b>=p[i].a){            c--;            p[a].b=max(p[i].b,p[a].b);        }        else        {            a++;            p[a].a=p[i].a;p[a].b=p[i].b;        }    }    return c;}int main(){    ios::sync_with_stdio(false);    cin.tie(0);    cin>>n>>m;int a,b;    int ans=0;    for(int i=1;i<=n;i++)    {        cin>>a>>b;s1[i].a=a,s1[i].b=b;    }    for(int i=1;i<=m;i++)    {        cin>>a>>b;s2[i].a=a;s2[i].b=b;    }    sort(s1+1,s1+n+1,cmp);    sort(s2+1,s2+m+1,cmp);    n=flow(s1,1,n);    m=flow(s2,1,m);    int f,g=1;int k=1;    for(int i=1;i<=n;i++)    {        int f=s1[i].a;        while(g<=m)        {            if(s2[g].b>=f)                break;            g++;        }        if(g>m){            while(i<=n)            {                ans+=s1[i].b-s1[i].a;                i++;            }            break;        }          a=s2[g].a;        ans+=min(max(0,a-f),s1[i].b-f);          b=s1[i].b;        while(g+1<=m&&s2[g+1].a<b)        {            ans+=s2[g+1].a-s2[g].b;            g++;        }        if(s2[g].b<b)            ans+=b-s2[g].b;    }    cout<<ans<<endl;    return 0;}