CodeForces-1341E Unexpected Guest （01BFS）

传送门：CodeForces-1341E

思路分析

这dis[i][j]表示表示到了第i个位置，距离上次等红灯后走了j秒所需要的最小等红绿灯次数
这就是一个01BFS的问题了，如果j=g的话就需要停留，次数就要+1，将新状态放到队尾，否则将新状态放到队首
没取出一个队首计算贡献，取min

样例输入

15 5
0 3 7 14 15
11 11

样例输出

45

AC代码

#include <functional>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <vector>
#include <string>
#include <cstdio>
#include <chrono>
#include <random>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N=1e4+10;

int n,m,r,g,a[N];
int dis[N][1010],vis[N][1010];

int bfs() {
    int ans=INF;
    deque<pair<int,int>>q;
    q.push_back(mp(0,0));
    vis[0][0]=1;
    while(!q.empty()) {
        auto u=q.front();
        q.pop_front();
        int pos=u.fi;
        int cost=u.se;
        int need=n-a[pos]+cost;
        if(need<=g) ans=min(ans,dis[pos][cost]*(r+g)+need);
        if(cost==g) {
            if(!vis[pos][0]) {
                dis[pos][0]=dis[pos][g]+1;
                vis[pos][0]=1;
                q.push_back(mp(pos,0));
            }
            continue;
        }
        if(pos>1) {
            int next=pos-1;
            int need=a[pos]-a[next]+cost;
            if(need<=g && !vis[next][need]) {
                dis[next][need]=dis[pos][cost];
                vis[next][need]=1;
                q.push_front(mp(next,need));
            }
        }
        if(pos<m) {
            int next=pos+1;
            int need=a[next]-a[pos]+cost;
            if(need<=g && !vis[next][need]) {
                dis[next][need]=dis[pos][cost];
                vis[next][need]=1;
                q.push_front(mp(next,need));
            }
        }
    }
    return ans==INF?-1:ans;
}

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    read(n,m);
    for(int i=1; i<=m; i++) read(a[i]);
    sort(a+1,a+1+m);
    read(g,r);
    printf("%d\n",bfs());




    return 0;
}
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