Codeforces Round #476 (Div. 2) [Thanks, Telegram!] D. Single-use Stones CF965D

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看到题目先想到这明显是一个最大流
但是边太多了不可能用模版做
已知有 最大流 = 最小割
那么最小割一定是割了连续l个点
所以直接找最小割就是答案

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
#include <assert.h>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ansn() printf("%d\n",ans)
#define lansn() printf("%lld\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pll pair<long long,long long>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const ll mod = 1000000009 ;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
//const ll infl = 100000000000000000;//1e17
const int maxn=  1e5+20;
const int maxm = 2e2+20;
//muv[i]=(p-(p/i))*muv[p%i]%p;
int in(int &ret) {
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-1:1;
    ret = (c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
    ret *=sgn;
    return 1;
}


int a[maxn];
int pre[maxn];
int main() {
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

    int w,l;
    cin>>w>>l;
    r1(i,w-1)sd(a[i]),pre[i] = pre[i-1] + a[i];
    int ans = inf;
    for(int i=l;i<w;++i)
        ans = min(ans,pre[i] - pre[i-l]);
    ansn();
    return 0;
}