codeforces-1131 (div2)

时间：2025-03-22 08:07:31

A.把右上角的凹缺口补上变成凸的就成了规则矩形

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int main(){

    LL w1,h1,w2,h2;

    scanf("%lld%lld%lld%lld",&w1,&h1,&w2,&h2);

    LL ans = w1 + w1 + h1 + h1 + h2 + h2 + ;

    Prl(ans);

    return ;

}

A

B.画在图上就是两条斜线之间有多少可以水平的线，显然是下面这条线的最高点和上面这条线的最低点作差，记得打一个vis标记记录哪些线取过了

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int main(){

    Sca(N);

    LL la = ,lb = ;

    LL ans = ;

    LL now = ;

    for(int i = ; i <= N ; i ++){

        LL a,b; scanf("%lld%lld",&a,&b);

        int t = max(max(la,lb),now);

        if(min(a,b) >= t){

            ans += min(a,b) - t + ;

            now = min(a,b) + ;

        }

        la = a; lb = b;

    }

    Prl(ans);

    return ;

}

B

C.开始觉得要二分，后来觉得要O(n3),仔细一看发现O(n)

显然取两条非递减的线，从最高点往两边降低，贪心的发现每次都升高最高点更低的那个路径就可以了。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn];

int b[maxn],c[maxn];

int main(){

    Sca(N);

    for(int i = ; i <= N ; i ++) Sca(a[i]);

    sort(a + ,a +  + N);

    int cnt1 = ,cnt2 = ;

    b[++cnt1] = a[];

    c[++cnt2] = a[];

    for(int i = ; i <= N ; i ++){

        if(b[cnt1] < c[cnt2]) b[++cnt1] = a[i];

        else c[++cnt2] = a[i];

    }

    for(int i = ; i <= cnt1; i ++) printf("%d ",b[i]);

    for(int i = cnt2; i >= ; i --) printf("%d ",c[i]);

    return ;

}

C

D.拓扑排序例题。

先把所有的等于号用并查集合并，找大小之间的冲突。然后用拓扑排序标记顺便判环。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

char MAP[maxn][maxn];

struct Edge{

    int to,next;

}edge[maxn * maxn * ];

int fa[maxn * ],tot,head[maxn * ],ind[maxn * ];

void init(){

    for(int i =  ; i <= N + M; i ++){

        head[i] = -;

        fa[i] = i;

        ind[i] = ;

    }

    tot = ;

}

void add(int u,int v){

    edge[tot].to = v;

    edge[tot].next = head[u];

    head[u] = tot++;

}

int find(int x){

    if(x == fa[x]) return x;

    return fa[x] = find(fa[x]);

}

void Union(int a,int b){

    a = find(a); b = find(b);

    fa[a] = b;

}

int ans[maxn];

int main(){

    Sca2(N,M); init();

    bool flag = ;

    for(int i = ; i <= N ; i ++){

        scanf("%s",MAP[i] + );

        for(int j = ; j <= M ; j ++){

            if(MAP[i][j] == '='){

                Union(i,j + N);

            }

        }

    }

    for(int i = ; i <= N && flag; i ++){

        for(int j = ; j <= M ; j ++){

            if(MAP[i][j] == '=') continue;

            int a = find(i),b = find(j + N);

            if(a == b){

                flag = ;

                break;

            }

            if(MAP[i][j] == '>'){

                add(b,a); ind[a]++;

            }else{

                add(a,b); ind[b]++;

            }

        }

    }

    if(!flag){

        puts("No");

        return ;

    }

    queue<int>Q;

    for(int i = ; i <= N + M; i ++){

        if(find(i) == i && !ind[i]){

            ans[i] = ;

            Q.push(i);

        }

    }

    while(!Q.empty()){

        int u = Q.front(); Q.pop();

        for(int i = head[u]; ~i; i = edge[i].next){

            int v = edge[i].to;

            ans[v] = max(ans[v],ans[u] + );

            ind[v]--;

            if(!ind[v]) Q.push(v);

        }

    }

    for(int i = ; i <= N + M; i ++){

        if(fa[i] == i && ind[i]) flag = ;

    }

    if(!flag){

        puts("No");

        return ;

    }

    puts("Yes");

    for(int i = ; i <= N ; i ++){

        printf("%d ",ans[find(i)]);

    }

    puts("");

    for(int j = N + ; j <= N + M; j ++){

        printf("%d ",ans[find(j)]);

    }

    return ;

}

D

E.dp[100000][30]记录到i这个字符串j的最长长度。

如果右边的字符串都为k，那么dp[i][j] = max(dp[i][j],len * (dp[i - 1][j] + 1) + dp[i - 1][j]);

否则字符就变为前后缀与他相同的最长长度相加

如果曾经出现过这个字符，这个字符的出现次数至少为1

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 1e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

LL ans[];

LL dp[maxn][];

LL pre[];

LL tmp[];

char str[];

int main(){

    Sca(N);

    for(int i = ; i <= N ; i ++){

        scanf("%s",str);

        LL len = strlen(str);

        LL a = ,b = len;

        for(int j =  ; j < ; j ++) tmp[j] = ;

        for(int j = ; j < len; j ++){

            if(j && str[j - ] == str[j]) pre[j] = pre[j - ] + ;

            else pre[j] = ;

            tmp[str[j] - 'a'] = max(tmp[str[j] - 'a'],pre[j]);

        }

        for(int j =  ; j < ; j ++){

            dp[i][j] = tmp[j];

            int l = ,r = len - ;

            while(l <= r && str[l] == j + 'a') l++;

            while(r > l && j + 'a' == str[r]) r--;

            if(l == len){

                dp[i][j] = max(dp[i][j],len * (dp[i - ][j] + ) + dp[i - ][j]);

            }else if(dp[i - ][j]){

                dp[i][j] = max(dp[i][j],l + len - r);

            }

        }

    }

    LL sum = ;

    for(int j =  ; j < ; j ++) sum = max(sum,dp[N][j]);

    Prl(sum);

    return ;

}

E

F.喜闻乐见的送分F题，开到就是赚到。

并查集模板题，加一个l[maxn],r[maxn]表示这个集合最左最右的数字

nxt[maxn]表示这个数字的下一个数字，每次匹配u,v都把u集合放在v集合的左边，更新一下几个数组

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int fa[maxn],r[maxn],l[maxn];

int nxt[maxn];

void init(){

    for(int i =  ; i <= N ; i ++){

        fa[i] = r[i] = l[i] = i;

        nxt[i] = -;

    }

}

int find(int x){

    if(x == fa[x]) return x;

    return fa[x] = find(fa[x]);

}

int main(){

    Sca(N);    init();

    for(int i = ; i <= N - ; i ++){

        int u,v; Sca2(u,v);

        u = find(u); v = find(v);

        nxt[r[u]] = l[v];

        l[v] = l[u];

        fa[u] = v;

    }

    int root = l[find()];

    for(int i = ; i <= N ; i ++){

        printf("%d ",root);

        root = nxt[root];

    }

    return ;

}

F

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