Submission #18105

#	Time	Username	Problem	Language	Result	Execution time	Memory
18105		tncks0121	도장 모으기 (JOI14_stamps)	C++14	0 / 100	38 ms	73948 KiB

stamps

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <string>
#include <functional>
#include <vector>
#include <deque>
#include <utility>
#include <bitset>
#include <limits.h>
#include <time.h>
#include <functional>
#include <numeric>
 
using namespace std;
typedef long long ll;
typedef unsigned long long llu;
typedef double lf;
typedef unsigned int uint;
typedef long double llf;
typedef pair<int, int> pii;

const int N_ = 3050;

int N;
ll T;
int U[N_], V[N_], D[N_], E[N_];

ll sum1[N_]; // U+V
ll sum2[N_]; // min(U+V, D+E)
ll C1[N_], C2[N_], C3[N_], C4[N_];

ll tb[N_][N_];
void upd (ll &t, ll v) {
	if(t > v) t = v;
}

int main() {
    scanf("%d%lld", &N, &T);
    for(int i = 1; i <= N; i++) scanf("%d%d%d%d", U+i, V+i, D+i, E+i);

    for(int i = 1; i <= N; i++) {
    	C1[i] = U[i] + V[i];
    	C2[i] = D[i] + E[i];
    	C3[i] = U[i] + E[i];
    	C4[i] = D[i] + V[i];
		for(int j = i-1; j > 0; j--) {
   			upd(C4[i], T * (i - j) * 2 + min(C1[i], C2[i]) + C4[j]);
    	}
    	sum1[i] = sum1[i-1] + U[i] + V[i];
    	sum2[i] = sum2[i-1] + min(U[i] + V[i], D[i] + E[i]);
    }

    for(int i = 1; i <= N; i++) {
    	for(int j = 1; j <= i; j++) tb[i][j] = (ll)1e18;
    }

	tb[1][1] = T + C1[1];
    for(int i = 1; i <= N; i++) {
    	for(int j = 1; j < i; j++) {
    		upd(tb[i+1][j], tb[i][j] - C3[i] + min(C1[i], C2[i]) + C3[i+1] + 2 * T);
    		upd(tb[i+1][i+1], tb[i][j] + C1[i+1] + T * (i-j+1));
    	}

    	{ // T[i][i]
    		upd(tb[i+1][i+1], tb[i][i] + C1[i+1] + T);
    		upd(tb[i+1][i], tb[i][i] - C1[i] + C3[i+1] + C4[i] + 2*T);
    	}
    }

    ll ans = (ll)1e18;
    for(int i = 1; i <= N; i++) {
    	for(int j = 1; j <= i; j++) {
    		ll val = tb[i][j] + sum1[N] - sum1[i] + T * (N+1 - j);
    		ans = min(ans, val);
    	}
    }

    printf("%lld\n", ans);

    return 0;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 75
• Subtask 3: 0 / 15