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题目链接:
Time Limit: 12000/6000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=2e5+20;
const int maxn=1e4+220;
const double eps=1e-12;
int n,m,w[N],a[N],b[N],p[N];
int x,y;
vector<int>st[N],en[N];
int findset(int x)
{
if(x==p[x])return x;
return p[x]=findset(p[x]);
}
int same(int x,int y)
{
int fx=findset(x),fy=findset(y);
if(fx>fy)p[fy]=fx;
else if(fx<fy)p[fx]=fy;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
For(i,1,n-1)read(w[i]);
int high=max(n,m)+10;
for(int i=0;i<=high;i++)
{
p[i]=i;
st[i].clear();
en[i].clear();
}
For(i,1,m)
{
read(a[i]);read(b[i]);
if(a[i]>b[i])swap(a[i],b[i]);
int r=a[i];
while(1)
{
r=findset(r);
if(r>=b[i])break;
same(r,r+1);
st[i].push_back(r);
}
}
for(int i=0;i<=high;i++)p[i]=i;
for(int i=m;i>0;i--)//找到第i天要关闭的线段
{
int r=a[i];
while(1)
{
r=findset(r);
if(r>=b[i])break;
same(r,r+1);
en[i].push_back(r);
}
}
int ans=0;
for(int i=1;i<=m;i++)
{
int len=st[i].size();
for(int j=0;j<len;j++)//加上这一天要打开的线段
{
int x=st[i][j];
ans+=w[x];
}
print(ans);
len=en[i].size();
for(int j=0;j<len;j++)//关闭这天需要关闭的线段;
{
int x=en[i][j];
ans-=w[x];
}
}
}
return 0;
}
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5785910.html