模拟赛2

模拟赛2

调侃一下这次模拟赛爆炸。。。

T1

链接:https://www.luogu.org/problemnew/show/U45562
模拟题.

/*我还是很喜欢你,像风走了八百里不问归期. */
#include 
#include 
#include 
#define rep(i,x,p) for(int i = x;i <= p;++ i)
const int maxN = 1000 + 7;

char s[maxN][maxN];
int L;
int a[maxN];

int main() {
    freopen("curse.in","r",stdin);
    freopen("curse.out","w",stdout);
    int n;
    scanf("%d",&n);
    rep(i,1,n)
        scanf("%s",s[i] + 1);
    L = strlen(s[1] + 1);
    rep(j ,1, L) {
        int num_1 = 0,num_0 = 0;
        rep(i , 1, n) 
            if(s[i][j] == '1') num_1 ++;
            else num_0 ++;
            if(num_1 > num_0) a[j] = 1;
    }
    rep(i,1,L) printf("%d", a[i]);
    return 0;
}

T2

神奇的题目,首先看到L满足单调性,二分.
然后剩下的就是如何处理红光和绿光如何使用了.
刚开始思路偏了,感觉是贪心,但是贪心的话,复杂度是\(n * log n\)的,这就有些说不过去了.
然后开始想DP.设状态的时候一直把法坛加入.复杂度一直是\(n^3\)的,而且无法优化.
一直想到比赛结束.当时我应该跳出思维.哎。。。。
状态还是比较容易想到.
因为如果\(R,G\)和的个数大于等于n的话.那么答案就是1.
所以R,G的状态数非常小.
设\(f[i][j]\)表示R使用了i个,G用了j个所能到达的最远处.
然后预处理所有的法坛,跳L步最多到哪,跳2*L步最多跳到哪
这道题就算完成了.

#include 
#include 
#include 
#include 
#define rep(i,x,p) for(int i = x;i <= p;++ i)
#define sep(i,x,p) for(int i = x;i >= p;-- i)
using namespace std;
const int maxN = 2000 + 7;

int a[maxN],n,R,G;
int f[maxN][maxN]; // f[i][j] 表示红色为i,绿色为j可以到达最远的地方.
int P[maxN],Q[maxN];

inline int read() {
    int x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
    return x * f;
}

bool calc(int L)
{
    memset(f, 0, sizeof(f));
    memset(P, 0, sizeof(P));
    memset(Q, 0, sizeof(Q));
    rep(i,1,n) {
        rep(j,i,n) {
            if(a[j] - a[i] + 1 <= L) P[i] = j;
            if(a[j] - a[i] + 1 <= 2 * L) Q[i] = j; 
        }
    }
    P[n + 1] = Q[n + 1] = n;
    rep(i,0,R) {
        rep(j,0,G) {
            if (i > 0) f[i][j] = max(f[i][j], P[f[i-1][j] + 1]);
            if (j > 0) f[i][j] = max(f[i][j], Q[f[i][j-1] + 1]);    
        }   
    }
    return f[R][G] == n;
}

int main() {
    n = read();R = read();G = read();
    rep(i,1,n) a[i] = read();
    sort(a + 1,a + n + 1);
    if(R + G >= n) {puts("1");return 0;}
    int l = 1,r = a[n] - a[1] + 1,ans;
    while(l <= r) {
        int mid = (l + r) >> 1;
        if(calc(mid)) {
            r = mid - 1;
            ans = mid;
        }
        else  l = mid + 1;
    }
    printf("%d",ans);
    return 0;
}

T3

K短路模板,不过我不会.
而且正解也不会证明.
从\(1\)跑一边最短路,从\(n\)跑一边最短路,枚举每一条边.
如果这条道路不跟1到n的最短路相通的话.那么就去最小值即可.

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define rep(i,x,p) for(int i = x;i <= p;++ i) 
#define sep(i,x,p) for(int i = x;i >= p;-- i)
const int maxN = 5000 + 7;
const int maxM = 200000 + 7;

int dis[maxN],dis2[maxN];
bool vis[maxN];

typedef pair  P;
priority_queue, greater
 > Q; 

struct Node {
    int v,nex,w;
}Map[maxM];
int num,head[maxN];

struct Node_1 {
    int u,v,w;
}C[maxM];

void add_Node(int u,int v,int w) {
    Map[++ num] = (Node) {v,head[u],w};
    head[u] = num;
    return;
}

void dij_1(int now) {
    memset(vis,0,sizeof(vis));
    memset(dis,0x3f,sizeof(dis));
    Q.push(make_pair(0,now));
    dis[now] = 0;
    while(!Q.empty()) {
        int u = Q.top().second;
        Q.pop();
        if(vis[u]) continue;
        vis[u] = 1;
        for(int i = head[u];i;i= Map[i].nex) {
            int v = Map[i].v;
            if(dis[u] + Map[i].w < dis[v]) {
                dis[v] = dis[u] + Map[i].w;
                Q.push(make_pair(dis[v],v));
            }
        }
    } 
    return ;
}

void dij_2(int now) {
    memset(vis,0,sizeof(vis));
    memset(dis2,0x3f,sizeof(dis2));
    Q.push(make_pair(0,now));
    dis2[now] = 0;
    while(!Q.empty()) {
        int u = Q.top().second;
        Q.pop();
        if(vis[u]) continue;
        vis[u] = 1;
        for(int i = head[u];i;i= Map[i].nex) {
            int v = Map[i].v;
            if(dis2[u] + Map[i].w < dis2[v]) {
                dis2[v] = dis2[u] + Map[i].w;
                Q.push(make_pair(dis2[v],v));
            }
        }
    } 
    return ;
}

inline int read() {
    int x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
    return x * f;
}

int main() {
    freopen("maze.in","r",stdin);
    freopen("maze.out","w",stdout);
    int n = read(),m = read() ;
    int u , v, w;
    rep(i,1,m) {
        u = read();v = read();w = read();
        add_Node(u,v,w);
        add_Node(v,u,w);
        C[i] = (Node_1) {u,v,w};
    }
    dij_1(1);
    dij_2(n);
    int tmp = dis[n];
    int ans = 0x3f3f3f3f;
    rep(i,1,m) {
        int u = C[i].u,v = C[i].v,w = C[i].w;
        if(dis[u] + dis2[v] + w != tmp) {
            ans = min(ans,dis[u] + dis2[v] + w);
        }
        if(dis2[u] + dis[v] + w != tmp) {
            ans = min(ans,dis2[u] + dis[v] + w);
        }
    }
    printf("%d",ans);
    return 0;
}

Last

这次比赛直接sb了,没什么好讲的.除了T2有价值,其余的全都是傻逼题.
放在PJ还可以.

转载于:https://www.cnblogs.com/tpgzy/p/9805103.html