Nothing for Nothing(十七)

题目：Subsegments

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5;

int a[maxn];
set<int>S;
multiset<int>M;
int main()
{
    int n, k;
    scanf("%d %d", &n, &k);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
    }
    for(int i = 1; i <= k; i++)
    {
        int v = a[i];
        if(!M.count(v))S.insert(v);
        else if(S.count(v))S.erase(v);
        M.insert(v);
    }
    if(S.size() == 0)printf("Nothing\n");
    else cout<<*(--S.end())<<endl;
    for(int i = k+1; i <= n; i++)
    {
        multiset<int>::iterator pos = M.find(a[i-k]);
        M.erase(pos);
        if(!M.count(a[i-k]) && S.count(a[i-k]))S.erase(a[i-k]);
        else if(M.count(a[i-k]) == 1 && !S.count(a[i-k]))S.insert(a[i-k]);
        if(!M.count(a[i]))S.insert(a[i]);
        else if(S.count(a[i]))S.erase(a[i]);
        if(S.size() == 0)printf("Nothing\n");
        else cout<<*(--S.end())<<endl;
        M.insert(a[i]);
    }

    return 0;
}

题目：Destroying Array

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
#define LL long long
LL a[maxn];
int p[maxn];
LL fa[maxn];
bool vis[maxn];
LL sum[maxn], ans[maxn];

inline int Find(int x)
{
	return fa[x]==x?x:fa[x]=Find(fa[x]);
}
inline void Init(int n){
       for(int i = 1; i <= n; i++)fa[i] = i;
}
int main(){
    int n;
    scanf("%d", &n);
    Init(n);
    for(int i = 1; i <= n; i++)scanf("%lld", &a[i]);
    for(int i = 1; i <= n; i++)scanf("%d", &p[i]);
    ans[n] = 0;
    for(int i = n; i > 1; i--){
        int t = p[i];
        sum[t] = a[t];
        if(vis[t-1]){
            int f1,f2;
            f1 = Find(t);
            f2 = Find(t-1);
            fa[f1] = f2;
            sum[f2] += sum[f1];
        }
        if(vis[t+1]){
            int f1,f2;
            f1 = Find(t);
            f2 = Find(t+1);
            fa[f1] = f2;
            sum[f2] += sum[f1];
        }
        ans[i-1] = max(sum[Find(t)], ans[i]);
        vis[t] = true;
    }
    for(int i = 1; i <= n; i++){
        printf("%lld\n", ans[i]);
    }

return 0;
}

题目：The Two Routes

思路：仔细读题意可知，从1~n之间不是存在一条直通的铁路就是公路，然后用最短路求另外一种就可以了。
代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;
#define Max 1<<20

const int Maxn = 410;

int ma1[Maxn][Maxn], dis[Maxn], ma2[Maxn][Maxn];
int vis[Maxn];
int n, m;
queue<int>q;
void SPFA(int st, int ma[Maxn][Maxn])
{
    int i, j;
    for(i = 1; i <= n; i++)
    {
        dis[i] = Max;
        vis[i] = 0;
    }
    vis[st] = 1;
    dis[st] = 0;
    q.push(st);
    while(!q.empty())
    {
        int t = q.front();
        vis[t] =  1;
        q.pop();
        for(i = 1; i <= n; i++)
        {
            if(dis[i] > dis[t] + ma[t][i])
            {
                dis[i] = dis[t] + ma[t][i];
                if(!vis[i])
                {
                    vis[i] = 1;
                    q.push(i);
                }
            }

        }
        vis[t] = 0;

    }

}

int main(){
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                ma1[i][j] = ma2[i][j] = Max;
            }
        }
        for(int i = 0; i < m; i++){
            int t1, t2;
            cin>>t1>>t2;
            ma1[t1][t2] = ma1[t2][t1] = 1;
        }
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                if(ma1[i][j] == Max){
                    ma2[i][j] = 1;
                }
            }
        }
        int ans = 0;
        if(ma1[1][n] == 1 || ma1[n][1] == 1){
            SPFA(1, ma2);
        }
        else {
            SPFA(1, ma1);
        }
        if(dis[n] == Max)printf("-1\n");
        else printf("%d\n", dis[n]);

return 0;
}

题目：Gargari and Bishops

思路：就是求出两个斜对角线，然后你会发现，只能是在横纵角标的和奇偶中选，然后贪心一下就可以了。
代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 2010;
int a[maxn][maxn];
LL r[maxn<<1], l[maxn<<1];

int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            scanf("%d", &a[i][j]);
            r[i+j] += a[i][j];
            l[n+i-j] += a[i][j];
        }
    }
    LL max1 = -1, max2 = -2;
    int x1, y1, x2, y2;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            LL ret = r[i+j]+l[n+i-j]-a[i][j];
            if(((i+j)&1) && ret > max1){
                max1 = ret;
                x1 = i;
                y1 =  j;
            }
            if(!((i+j)&1) && ret > max2 ){
                max2 = ret;
                x2 = i;
                y2 = j;
            }
        }
    }
    printf("%lld\n", max1+max2);
    printf("%d %d %d %d\n", x1, y1, x2, y2);

return 0;
}

题目：Infinite Maze

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 2000;
char mp[maxn][maxn];
int n,m;

typedef pair<int,int> pII;
const int dx[4]={1,-1,0,0};
const int dy[4]={0,0,1,-1};

pII last[maxn][maxn];
bool ans=false;

void dfs(int x,int y)
{
	int mx=x%n,my=y%m;
	if(mp[mx][my]=='#'||ans||last[mx][my]==pII(x,y)) return;
	if(last[mx][my] != pII(0,0))
	{
		ans=true; return ;
	}
	last[mx][my]=pII(x,y);
	for(int i=0;i<4;i++)
	{
		dfs(x+dx[i],y+dy[i]);
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
		scanf("%s",mp[i]);
	int st,en;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			if(mp[i][j]=='S')
				st=i,en=j;
		}
	}
	dfs(st+n*m,en+n*m);
	if(ans) puts("Yes");
	else puts("No");
	return 0;
}