4.17 模拟总结

昨天熬夜了早上起来有点迷糊

整体就那样吧

上节课（4.14）开始算走了4天了

第一题

第一遍开始以为是规律题或者说结论题（ACM黑匣子之后每道字符串题都像结论题淦）

拿大样例推了半个小时 推了个大概 估计能拿40pts 结果真就45pts

10
tltltttlttltlttltttttlttlttlttltltlttlt
tltltlttlttttltltttttlttltltlttltlttlt
ttlttltlttltlttltttltltltltltttlttltlt
ttttltttltttlttttttttttlttltltltlttltt
ltttltttttttltttltlttttttltltttlttlttt
tllltllltltttlltlltlltltltllttlllllllll
llltllllttltlllttltllltlltllttlltllllll
tttltttltltltttlttttltttttltttlttlttlt
lltllttlllllttltllltllltltllllltttttllt
tltlttltltltttltltltttlttttltltltltltla

0 1
0 1
0 1
0 1
1 1
0 0
0 0
0 1
0 0
0 0

先看0 0

0 0几大特点：

1、字符串含有非"t l "元素 必0 0

2、任意两个l相连 必0 0

对于其他的：

3、开头是t 预处理tui = 1， 开头是l 预处理lao = 1；

就这么暴力

原代码 45pts

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <cmath>
//#define int long long
using namespace std;
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;
}
string s;
bool t = 0;
bool l = 0;
int lao = 1;
int tui = 1;
int main(){
    freopen("laotui.in","r",stdin);
    freopen("laotui.out","w",stdout);
    int T = read();
    while(T --){
        lao = 1;
        tui = 1;
        t = 0;
        l = 0;
        cin>>s;
        if(s[0] == 't') tui = 1,t = 1,lao = 0;
        if(s[0] == 'l') lao = 1,l = 1;
        for(register int i = 1;i <= s.size() - 1; i ++){
            if(l && s[i] == 'l'){
                printf("0 0\n");
                break;
            }
            if(s[i] != 't' && s[i] != 'l'){
                printf("0 0\n");
                break;
            }
            if(!t && s[i] == 't') t = 1,l = 0;
            if(!l && s[i] == 'l') l = 1,t = 0;
            if(i == s.size() - 1){
                printf("%d %d\n",lao,tui);
            }
        }
    }
    return 0;
}

题解：

第二题：

上手就是dijiska

显然不对

因为没考虑闪现

看小样例

先把0 2 100加速

然后再dijiska

小样例解决；

那就是 先贪心路程最大的加速掉 然后跑最短路

大样例不对

发现自己写的有向最短

改成无向最短

本来想着是先dij再加速 后来发现这样得写记录最短路径 复杂度撕不开 而且小样例也不对

于是过大样例

结果40pts

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <cmath>
//#define int long long
using namespace std;
const int M = 1e4 + 5;
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;
}
const int MaxM = 5e4 + 10;
struct edge{
    int to, dis, next,num;
}e[MaxM];
int head[MaxM];
int dis[MaxM];
bool vis[MaxM];
int cnt = 0;
int n;
int m;
int s;
int k;
inline void add_edge( int u, int v, int d ) {
    cnt++;
    e[cnt].dis = d;
    e[cnt].to = v;
    e[cnt].next = head[u];
    e[cnt].num = cnt;
    head[u] = cnt;
}
bool cmp(edge a,edge b) {
    return a.dis > b.dis;
}
bool Cmp(edge a,edge b){
    return a.num < b.num;
}
struct node{
    int dis;
    int pos;
    bool operator <( const node &x )const{
        return x.dis < dis;
    }
};
priority_queue<node> q;
inline void dijkstra() {
    dis[s] = 0;
    q.push((node) {0, s});
    while (!q.empty()) {
        node tmp = q.top();
        q.pop();
        int x = tmp.pos;
        if (vis[x]) continue;
        vis[x] = 1;
        for (register int i(head[x]); i; i = e[i].next) {
            int y = e[i].to;
            if (dis[y] > dis[x] + e[i].dis) {
                dis[y] = dis[x] + e[i].dis;
                if (!vis[y]) q.push((node) {dis[y], y});
            }
        }
    }
}
int main() {
    freopen("overwatch.in", "r", stdin);
    freopen("overwatch.out", "w", stdout);
    n = read();
    m = read();
    k = read();
    s = 0;
    fill(dis + 1, dis + 1 + n, 0x7fffffff);
    for (register int i(0); i <= m - 1; ++i) {
        register int u = read();
        register int v = read();
        register int d = read();
        add_edge(u, v, d);
        add_edge(v, u, d);
    }
    if (k != 0) {
        sort(e + 1, e + 1 + cnt, cmp);
        for (register int i = 1; i <= k; i++) {
            e[i].dis = 0;
        }
        sort(e + 1, e + 1 + cnt, Cmp);
    }
    dijkstra();
    cout << dis[n - 1] << endl;
    return 0;
}

玩过守望先锋的估计会写过吧（bushi

题解：

第三题

好几年前玩过皇室战争（雾

一眼区间dp

时间不够了（20分钟）

转移方程瞎推了一波

最开始想两组为一个单位前后选着算

写着写着就成了01背包

小样例不过

把dp打开看 一个正确答案都没有

最后瞎处理了一波

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <cmath>
#define int long long
using namespace std;
const int M = 1e4 + 5;
inline int read() {
    int x = 0, dp = 1;
    char c = getchar()；
    while (c < '0' || c > '9') {
        if (c == '-') dp = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * dp;
}
int n;
int k;
int dp[M][M];
int c[M];
int e[M];
struct A{
        int w;
        int v;
}a[M];
signed main() {
    freopen("clash.in", "r", stdin);
    freopen("clash.out", "w", stdout);
        n = read();
        k = read();
        for (register int i(1); i <= n; i++) {
            a[i].w = read();
            a[i].v = read();
            c[i] = c[i - 1] + a[i].v;
        }
        for (register int i(1); i <= n - 1; i++) {
            if (a[i].w + a[i + 1].w <= k) {
                dp[i][i + 1] = a[i].v + a[i + 1].v;
            }
        }
        for (register int l(2); l <= n; l += 2) {
            for (register int i(1); i <= n + 1 - l; i++) {
                if (a[i].w + a[i + l - 1].w <= k) dp[i][i + l - 1] = max(dp[i][i + l - 1],dp[i + 1][i + l - 2] + a[i].v +a[i + l - 1].v);
                for (register int k(i + 1); k <= i + l - 3; k += 2) dp[i][i + l - 1] = max(dp[i][i + l - 1], dp[i][k] + dp[k + 1][i + l - 1]);
            }
        }
        for (register int i(2); i <= n; i++) {
            e[i] = e[i - 1];
            for (register int j = i % 2; j <= i - 2; j += 2) {
                e[i] = max(e[i], e[j] + dp[j + 1][i]);
            }
        }
        cout << e[n] << endl;
        return 0;
    }

果然0分

题解 ： 

整体那样吧 图论-bot dp-bot还需要再看一看 

就这样吧