A. Attack
poj 3123
板子题 斯坦纳树
板子 如下
const int maxm=2009;
const int maxn=39;
const int inf=0x3f3f3f3f;
struct Edge{
int v,w,next;
}edge[maxm];
int head[maxn],top;
void init(){
memset(head,-1,sizeof(head));
top=0;
}
void add(int u,int v,int w){
edge[top].v=v;
edge[top].w=w;
edge[top].next=head[u];
head[u]=top++;
}
queue<int> q;
int stns[1<<9][maxn];
bool vis[1<<9][maxn];
void spfa(int sta){
int u,v,w;
while(!q.empty()){
u=q.front(); q.pop();
vis[sta][u]=0;
for(int i=head[u];i!=-1;i=edge[i].next){
v=edge[i].v;
w=edge[i].w;
if(stns[sta][v]>stns[sta][u]+w){
stns[sta][v]=stns[sta][u]+w;
if(!vis[sta][v]){
vis[sta][v]=1;
q.push(v);
}
}
}
}
}
int can[maxn];//为1表示这个点是点集中的点;
int STNS(int n,int id){//所有点数,斯坦纳点个数;
int MAX=(1<<id)-1;
memset(stns,0x3f,sizeof(stns));
for(int i=1;i<=n;++i)
if(can[i]) stns[(1<<id)-1][i]=0;
for(int sta=1;sta<=MAX;++sta){
for(int i=1;i<=n;++i){
for(int son=sta&(sta-1);son;son=(son-1)&sta)
stns[sta][i]=min(stns[sta][i],stns[son][i]+stns[sta-son][i]);
if(stns[sta][i]!=inf&&vis[sta][i]==0){
q.push(i);
vis[sta][i]=1;
}
}
spfa(sta);
}
int res=inf;
for(int i=1;i<=n;++i)
res=min(res,stns[MAX][i]);
return res;
}
F. Sequence
找规律 偶数长度 贡献度 0
奇数长度 奇数位贡献 1次
#include <iostream>
#include <cstdio>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int data[maxn];
int tree[maxn << 2][2];
void push_up(int rt) {
tree[rt][0] = tree[rt << 1][0] ^ tree[rt << 1 | 1][0] ;
tree[rt][1] = tree[rt << 1][1] ^ tree[rt << 1 | 1][1] ;
}
void buid(int l, int r, int rt) {
if(l == r) {
if(l % 2 == 0)
tree[rt][0] = data[l];
else
tree[rt][1] = data[l];
return;
}
int mid = l + r >> 1;
buid(l, mid, rt << 1);
buid(mid + 1, r, rt << 1 | 1);
push_up(rt);
}
void update(int L, int d, int l, int r, int rt, int k) {
if(l == r) {
tree[rt][k] = d;
return;
}
int mid = l + r >> 1;
if(L <= mid)
update(L, d, l, mid, rt << 1, k);
else
update(L, d, mid + 1, r, rt << 1 | 1, k);
push_up(rt);
}
int find(int L, int R, int l, int r, int rt, int k) {
if(L <= l && R >= r) {
return tree[rt][k];
}
int mid = (l + r) >> 1;
ll ans = 0;
if(L <= mid)
ans ^= find(L, R, l, mid, rt << 1, k);
if(R > mid)
ans ^= find(L, R, mid + 1, r, rt << 1 | 1, k);
return ans;
}
int main() {
int t;
scanf("%d", &t);
int cas = 1;
while(t --) {
int n, m;
scanf("%d %d", &n, &m);
for(int i = 1; i <= n; i++) scanf("%d", &data[i]);
buid(1, n, 1);
printf("Case #%d:\n", cas++);
while(m--) {
int cmd, l, r;
scanf("%d %d %d", &cmd, &l, &r);
if(cmd == 0)
if(l % 2 == 0)
update(l, r, 1, n, 1, 0);
else
update(l, r, 1, n, 1, 1);
else if((r - l + 1) % 2 == 0) printf("0\n");
else if(l % 2 == 0) printf("%d\n", find(l, r, 1, n, 1, 0));
else printf("%d\n", find(l, r, 1, n, 1, 1));
}
}
return 0;
}
G. Winner
我们按3个排序 把最大的标记1 然后权值有小到大 不断 扫 把一之后跟新 为1 直到没有跟新退出
#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct node {
int num;
int val;
bool operator < (const node &a) const {
return val < a.val;
}
} data[maxn], data2[maxn], data3[maxn];
bool vis[maxn];
int main() {
int n, m;
scanf("%d %d", &n, &m);
int a = 0, b = 0, c = 0;
for(int i = 1; i <= n; i++) {
data[i].num = i;
scanf("%d", &data[i].val);
if(data[i].val > data[a].val) a = i;
}
for(int i = 1; i <= n; i++) {
data2[i].num = i;
scanf("%d", &data2[i].val);
if(data2[i].val > data2[b].val) b = i;
}
for(int i = 1; i <= n; i++) {
data3[i].num = i;
scanf("%d", &data3[i].val);
if(data3[i].val > data3[c].val) c = i;
}
vis[a] = vis[b] = vis[c] = 1;
sort(data + 1, data + 1 + n);
sort(data2 + 1, data2 + 1 + n);
sort(data3 + 1, data3 + 1 + n);
while(1) {
bool flag = 0;
bool tt = 0;
for(int i = 1; i <= n; i++) {
if(vis[data[i].num] == 1) flag = 1;
if(flag == 1 && vis[data[i].num] == 0) {
tt = 1;
vis[data[i].num] = 1;
}
}
flag = 0;
for(int i = 1; i <= n; i++) {
if(vis[data2[i].num] == 1) flag = 1;
if(flag == 1 && vis[data2[i].num] == 0) {
tt = 1;
vis[data2[i].num] = 1;
}
}
flag = 0;
for(int i = 1; i <= n; i++) {
if(vis[data3[i].num] == 1) flag = 1;
if(flag == 1 && vis[data3[i].num] == 0) {
tt = 1;
vis[data3[i].num] = 1;
}
}
if(tt == 0) break;
}
while(m--) {
int p;
scanf("%d", &p);
if(vis[p] == 1) printf("YES\n");
else printf("NO\n");
}
return 0;
}