以后WA了T了看数组；
暑假四次数组下标超界，多次数组开小，暂时没有访问到负下标
题解；
第一题；这道题可以转换为颜色相同的点缩成一个点，每次可以将两个点合并成同一点，问最少几次将所有点合并成一个点；
开始想到并查集+贪心合并度数最多的并查集，但这样是有问题的，比如度数一样时，选择的先后顺序是有影响的；
正解：缩点+找直径，如果是一条黑白相间的链，就是点数/2, 而树上任何一条直径都会有一个点经过直径，我们从交点开始往外延伸，发现最长延伸就是直径本身；
思想：从特殊到一般
#include<bits/stdc++.h>
using namespace std;
const int M = 1e5 + 10;
#define ex(i, u) for(int i = h[u]; i; i = G[i].nxt)
int h[M], col[M], tot, fa[M], rec[M], vec[M], dep[M];
struct edge{int nxt, v;}G[M * 2];
void add(int u, int v){
G[++tot].nxt = h[u]; h[u] = tot; G[tot].v = v;
}
int find(int u){
if(u == fa[u])return u;
return fa[u] = find(fa[u]);
}
void uni(int u, int v){
fa[find(u)] = find(v);
}
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 rt, res;
void dfs(int u, int f){
//fprintf(stderr , "%d %d\n", u, f);
dep[u] = dep[f] + 1;
ex(i, u){
int v = G[i].v;
if(v == f)continue;
dfs(v, u);
}
if(dep[u] > dep[res])res = u;
}
int main(){
freopen("color.in","r",stdin);
freopen("color.out","w",stdout);
int T;
scanf("%d", &T);
while(T--){
int n = read();
for(int i = 1; i <= n; i++)
col[i] = read(), h[i] = 0, fa[i] = i;
tot = 0;
int u, v, p = 0;
for(int i = 1; i < n; i++){
u = read(), v = read();
if(col[u] == col[v])uni(u, v);
else vec[++p] = u, rec[p] = v;
}
for(int i = 1; i <= p; i++){
int x = find(rec[i]), y = find(vec[i]);
add(x, y), add(y, x); //fprintf(stderr, "%d %d\n", y, x);
}
for(int i = 1; i <= n; i++)
if(i == find(i)){
res = rt = 0;
memset(dep, 0, sizeof(dep));
dfs(i, 0);
memset(dep, 0, sizeof(dep));
rt = res;
dfs(rt, res = 0);
break;
}
printf("%d\n", dep[res]/ 2);
}
return 0;
}
第二题：模拟，我们发现很多路径是重复的，所以要想办法避免；又发现格子最多有150^2个，所以我们只需要记录格子是否被访问；每次从一个格子暴力沿八个方向走，如果到一个格子也有相同的方向，如果该格子走的远，就停下来，让这个格子完成任务，如果该格子走的不如当前远，那么他的延伸就是废的，去掉；
每个格子最多朝8个方向时被重复，就是O(300*300*8);
这道题我数组开的N，循环的时候也循环到N， 开了O2全WA，都学这么久了，怎么还在犯这种低级错误！！！
而且上次考试又是少写了一个等号全WA！！！该好好练练码力了；
#include<bits/stdc++.h>
using namespace std;
const int N = 306;
int zl[10][2] = {{0, 0}, {0, 1}, {1, 1}, {1, 0}, {1, -1}, {0, -1}, {-1, -1}, {-1, 0}, {-1, 1}};
int go[10][2] = {{0, 0}, {2, 8}, {1, 3}, {2, 4}, {3, 5}, {4, 6}, {5, 7}, {6, 8}, {1, 7}};
int n;
int mp[N+11][N+11][10], a[50];
bool vis[N+11][N+11], done[N+11][N+11];
struct node{int x,y,dep,fx;};
void bfs(){
queue <node> q;
q.push((node){152, 152, 1, 1});
mp[152][152][1] = 1;
while(!q.empty()){
node u = q.front(); q.pop();
done[u.x][u.y] = 1;
int i = u.fx;
int x = u.x + zl[i][0] * a[u.dep], y = u.y + zl[i][1] * a[u.dep];
if(a[mp[x][y][go[i][0]]] < a[u.dep + 1]){
mp[x][y][go[i][0]] = u.dep + 1;q.push((node){x, y, u.dep + 1, go[i][0]});
}
if(a[mp[x][y][go[i][1]]] < a[u.dep + 1]){
mp[x][y][go[i][1]] = u.dep + 1;q.push((node){x, y, u.dep + 1, go[i][1]});
}
}
}
void work(){
for(int i = 1; i <= N; i++)
for(int j = 1; j <= N; j++){
if(!done[i][j])continue;
for(int k = 1; k <= 8; k++){
if(mp[i][j][k]){
//printf("%d %d %d %d\n", i, j, k, mp[i][j][k]);
int x = i, y = j;
for(int z = 1; z <= a[mp[i][j][k]]; z++){
x += zl[k][0], y += zl[k][1];
if(a[mp[x][y][k]] > a[mp[i][j][k]] - z)break;
else mp[x][y][k] = 0;
vis[x][y] = 1;
}
vis[i][j] = 1;
}
}
}
}
int main(){
freopen("grow.in","r",stdin);
freopen("grow.out","w",stdout);
//int cc=clock();
int ans = 0;
scanf("%d", &n);
for(int i = 1; i <= n; i++)scanf("%d", &a[i]);
a[1]--;
bfs();
work();
for(int i = 1; i <= N; i++)
for(int j = 1; j <= N; j++)if(vis[i][j])ans++;//,printf("%d %d\n",i,j);
printf("%d\n", ans);
//int tt=clock();
//cout<<tt-cc;
}
第三题：dp,定义状态 ：
0 1 2 3 4
# 2 20 201 2017
dp[s][t]表示从状态s转移到状态t的最少删除字符，我们枚举中间状态r, dp[s][t] = min(dp[s][r] + dp[r][t]);
但不可能每次询问都做一次，所以用线段树做，转移相当于线段树合并；代码可以学习一下
#include<bits/stdc++.h>
using namespace std;
const int inf = 1e8, M = 1e5 + 10;
int n;
char s[M];
struct Info{//没有放在Node里是因为重载运算符不能返回一个指针
int tr[5][5];
// 0 1 2 3 4
// $ 2 20 201 2017
void init(int v){
memset(tr, 0x3f, sizeof(tr));
if(v == 3 || v == 4 || v == 5 || v == 8 || v == 9){
for(int i = 0; i < 5; i ++)tr[i][i] = 0;
}
if(v == 2){
tr[0][0] = 1;
tr[0][1] = 0;
tr[1][1] = tr[2][2] = tr[3][3] = tr[4][4] = 0;
}
if(v == 0){
tr[1][2] = 0;
tr[1][1] = 1;
tr[0][0] = tr[2][2] = tr[3][3] = tr[4][4] = 0;
}
if(v == 1){
tr[2][3] = 0;
tr[2][2] = 1;
tr[0][0] = tr[3][3] = tr[4][4] = tr[1][1] = 0;
}
if(v == 7){
tr[3][3] = 1;
tr[3][4] = 0;
tr[1][1] = tr[4][4] = tr[2][2] = tr[0][0] = 0;
}
if(v == 6){
tr[3][3] = tr[4][4] = 1;
tr[0][0] = tr[1][1] = tr[2][2] = 0;
}
}
};
struct Node{
Info info;
Node *ls, *rs;
}pool[(M << 2)], *tail = pool, *root;
Info operator + (const Info &a, const Info &b){//重载运算符写法
Info rt;
memset(&rt, 0x3f, sizeof(rt));//取地址
for(int i = 0; i < 5; i++)
for(int j = 0; j < 5; j++)
for(int k = i; k <= j; k++)
rt.tr[i][j] = min(rt.tr[i][j], a.tr[i][k] + b.tr[k][j]);
/*for(int i = 0;i<5;i++)for(int j=0; j<5;j++)
printf("%d %d %d\n",i,j,zero->tr[i][j]);puts("");*/
return rt;
}
Node * build(int lf = 1, int rg = n){
Node *nd = ++tail;
if(lf == rg)nd->info.init(s[lf - 1] - ‘0‘);
else {
int mid = (lf + rg) >> 1;
nd->ls = build(lf, mid);
nd->rs = build(mid + 1, rg);
nd->info = nd->ls->info + nd->rs->info;
}
return nd;
}
Info query(int L, int R, Node * nd = root, int lf = 1, int rg = n){
if(lf >= L && rg <= R)return nd->info;
int mid = (lf + rg) >> 1;
if(L > mid)return query(L, R, nd->rs, mid + 1, rg);
else if(R <= mid)return query(L, R, nd->ls, lf, mid);
else return query(L, R, nd->ls, lf, mid) + query(L, R, nd->rs, mid + 1, rg);
}
int Query(int l, int r){
Info ans = query(l, r);
return ans.tr[0][4] < inf ? ans.tr[0][4] : -1;
}
int main(){
freopen("year.in","r",stdin);
freopen("year.out","w",stdout);
scanf("%s", s);
n = strlen(s);
root = build();
int q, l, r;
scanf("%d", &q);
while(q--){
scanf("%d%d", &l, &r);
printf("%d\n", Query(l, r));
}
}
原文地址：https://www.cnblogs.com/EdSheeran/p/9538252.html
时间： 2024-10-11 00:22:26