Codeforces 311C Fetch the Treasure 取模意义下的最短路 (看题解)

Fetch the Treasure

感觉这题很nb啊， 虽然套了一个一点都不有趣的壳子。

我们注意到 k 的值在 1e4以内， 我们用d[ i ] 表示在模 k == i 能达到的最小的值是谁。

用最短路取更新。。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);
using namespace std;
const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);
template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}
LL h;
int n, m, k;
LL a[N];
int c[N];
LL d[N];
bool ban[N];
vector<PII> vc[N];
set<PII> Set;
int main() {
memset(d, INF, sizeof(d));
scanf("%lld%d%d%d", &h, &n, &m, &k);
d[0] = k;
for(int i = 1; i <= n; i++) scanf("%lld%d", &a[i], &c[i]), a[i]--;
for(int i = 1; i <= n; i++) {
if(a[i] % k == 0) {
Set.insert(mk(c[i], -i));
ban[i] = true;
}
}
while(m--) {
int op; scanf("%d", &op);
if(op == 1) {
LL x; scanf("%lld", &x);
priority_queue<PLI, vector<PLI>, greater<PLI> > que;
chkmin(d[x % k], x);
for(int i = 0; i < k; i++) if(d[i] < INF) que.push(mk(d[i], i));
while(!que.empty()) {
int u = que.top().se; LL dis = que.top().fi;
que.pop();
if(dis > d[u]) continue;
int v = (u + x) % k;
if(chkmin(d[v], dis + x)) {
que.push(mk(d[v], v));
}
}
for(int i = 1; i <= n; i++) {
if(ban[i]) continue;
if(d[a[i] % k] > a[i]) continue;
ban[i] = true;
Set.insert(mk(c[i], -i));
}
} else if(op == 2) {
int x, y;
scanf("%d%d", &x, &y);
if(ban[x]) {
Set.erase(mk(c[x], -x));
c[x] -= y;
Set.insert(mk(c[x], -x));
} else {
c[x] -= y;
}
} else {
if(!SZ(Set)) puts("0");
else {
printf("%dn", Set.rbegin()->fi);
Set.erase(*Set.rbegin());
}
}
}
return 0;
}
/*
*/