NOI2014
DAY 1
起床困难综合症
没怎么想正解，打了一个70分暴力
考虑到OR与OR之间的运算顺序对答案没有影响，即 ( ans OR x ) OR y 与 ( ans OR y ) OR x 相等，可以将多个连续的OR运算合并成一个，AND，XOR 同理
暴力枚举初始攻击力 0 ~ m
又存在多个数 OR/XOR/AND x 答案相等，如 5 OR 10 = 15, 6 OR 10 = 15, 13 OR 10 = 15，可记忆化
#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
inline int read()
{
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
return x * f;
}
int n, m, num;
set<pair<int, int> >st;
struct node
{
char op; int x;
node(){}
node(char o, int xx) {op = o; x = xx;}
}q[N];
int main()
{
n = read(); m = read();
for(int i = 1; i <= n; i++)
{
char s[5]; scanf("%s", s);
if(s[0] != q[num].op) q[++num] = node(s[0], read());
else
{
int x = read();
if(q[num].op == 'O') q[num].x |= x;
else if(q[num].op == 'X') q[num].x ^= x;
else q[num].x &= x;
}
if(q[num].op == 'A' && !q[num].op)
{
num = 1;
q[num] = node('A', 0);
}
}
int ans = 0;
for(int x = 0; x <= m; x++)
{
int t = x; bool fail = false;
for(int i = 1; i <= num; i++)
{
if(q[i].op == 'O') t |= q[i].x;
else if(q[i].op == 'X') t ^= q[i].x;
else t &= q[i].x;
if(st.find(make_pair(i, t)) == st.end()) st.insert(make_pair(i, t));
else {fail = true; break;}
}
if(!fail) ans = max(ans, t);
}
cout << ans;
return 0;
}
emmm......
然后又想了一下正解，也很简单啊，早知道就不去耗第三题了
单独考虑二进制下的每一位，分别计算初始值选择 0 / 1 得到的最终答案
初始值有上限，所以当选 0 / 1 效果相同时选 0，仅当选 1 效果更优且不会使值大于 m 时选 1
#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
inline int read()
{
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
return x * f;
}
int n, m, num;
set<pair<int, int> >st;
struct node
{
char op; int x;
node(){}
node(char o, int xx) {op = o; x = xx;}
}q[N];
int deal(int k, int i, int x)
{
if(k != -1)
{
int t = (q[i].x & (1 << k))? 1:0;
if(q[i].op == 'O') return x | t;
if(q[i].op == 'X') return x ^ t;
if(q[i].op == 'A') return x & t;
}
else
{
if(q[i].op == 'O') return x | q[i].x;
if(q[i].op == 'X') return x ^ q[i].x;
if(q[i].op == 'A') return x & q[i].x;
}
}
int main()
{
n = read(); m = read();
for(int i = 1; i <= n; i++)
{
char s[5]; scanf("%s", s);
if(s[0] != q[num].op) q[++num] = node(s[0], read());
else
{
int x = read();
if(q[num].op == 'O') q[num].x |= x;
else if(q[num].op == 'X') q[num].x ^= x;
else q[num].x &= x;
}
if(q[num].op == 'A' && !q[num].op)
{
num = 1;
q[num] = node('A', 0);
}
}
int ans = 0;
for(int k = 30; k >= 0; k--)
{
if((1 << k) > m) continue;
int x[2] = {0, 1};
for(int i = 1; i <= num; i++)
{
x[0] = deal(k, i, x[0]);
x[1] = deal(k, i, x[1]);
}
if(x[1] && !x[0])
if(ans + (1 << k) <= m) ans += (1 << k);
}
for(int i = 1; i <= num; i++) ans = deal(-1, i, ans);
cout << ans;
return 0;
}
魔法森林
数据有点水啊，二分 + dfs有20分，竟然过掉了一个500,3000的点
#include <bits/stdc++.h>
using namespace std;
const int N = 50005;
const int M = 200005;
inline int read()
{
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
return x * f;
}
int n, m, ax, bx;
int num, tmp1[M], to[M], tmp2[M], va[M], tmp3[M], vb[M], tmp4[M], nt[M], tmp5[M], p[N];
void add(int x, int y, int a, int b) {to[++num] = y; va[num] = a; vb[num] = b; nt[num] = p[x]; p[x] = num;}
bool tmp[M], vis[N];
bool dfs(int x, int a, int b, int mx)
{
if(a + b > mx) return false;
if(x == n) return true;
bool f = false;
for(int e = p[x]; e; e = nt[e])
{
if(vis[to[e]]) continue;
vis[to[e]] = true;
if(dfs(to[e], max(a, va[e]), max(b, vb[e]), mx)) return true;
vis[to[e]] = false;
}
return false;
}
int main()
{
n = read(); m = read();
for(int i = 1; i <= m; i++)
{
int x = read(), y = read(), a = read(), b = read();
if(x == y) continue;
add(x, y, a, b); add(y, x, a, b);
ax = max(ax, a); bx = max(bx, b);
}
int l = 0, r = ax + bx, ans = -1;
while(l <= r)
{
int mid = (l + r) >> 1;
memset(vis, false, sizeof(vis)); vis[1] = true;
if(dfs(1, 0, 0, mid)) {ans = mid; r = mid - 1;}
else l = mid + 1;
}
cout << ans;
return 0;
}
若只存在一个权值，即求最大边权最小，直接 spfa 即可
对于两种权值限制，可先对a进行排序，依次加边并对 b 跑 spfa，ans = max( ai + dis[n] );
每次添加的边仅对两端点有影响，不必重跑 spfa
#include <bits/stdc++.h>
using namespace std;
const int N = 50005;
const int M = 200005;
inline int read()
{
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
return x * f;
}
int n, m;
struct node
{
int x, y, a, b;
node(){}
node(int xx, int yy, int aa, int bb) {x = xx; y = yy; a = aa; b = bb;}
}q[M];
bool cmp(node A, node B) {return A.a < B.a;}
int num, to[M], w[M], nt[M], p[N];
void add(int x, int y, int v) {to[++num] = y; w[num] = v; nt[num] = p[x]; p[x] = num;}
int dis[N]; bool vis[N]; queue<int> Q;
void spfa()
{
while(!Q.empty())
{
int k = Q.front(); Q.pop(); vis[k] = false;
for(int e = p[k]; e; e = nt[e])
{
if(dis[to[e]] > max(dis[k], w[e]))
{
dis[to[e]] = max(dis[k], w[e]);
if(!vis[to[e]]) {vis[to[e]] = true; Q.push(to[e]);}
}
}
}
}
int main()
{
n = read(); m = read();
for(int i = 1; i <= m; i++)
{
q[i].x = read(); q[i].y = read(); q[i].a = read(); q[i].b = read();
}
sort(q + 1, q + 1 + m, cmp);
memset(dis, 63, sizeof(dis)); dis[1] = 0;
int ans = 1e8;
for(int i = 1; i <= m; i++)
{
add(q[i].x, q[i].y, q[i].b); add(q[i].y, q[i].x, q[i].b);
Q.push(q[i].x); Q.push(q[i].y); vis[q[i].x] = vis[q[i].y] = true;
spfa();
ans = min(ans, dis[n] + q[i].a);
}
if(ans == 1e8) cout << -1; else cout << ans;
return 0;
}
显然复杂度太玄学过不了
哎呀，好尴尬
显然复杂度太玄学过不掉
再次考虑先前的操作，对 b 求经过的最小的最大边权，可用最小生成树求解
按 a 从小到大依次加边，即维护动态最小生成树，lct
还没写完。。。
消除游戏
太麻烦啦，不想写
day1小结
第一次提交：70 + 10 + 0
不看题解进行修改：100 + 20 + 0
emmmmm，NOI要真考成这要得出大事。。。
