TJOI2013 DAY2
第一题：明显先处理出最终序列，然后用线段树求解。处理最终序列可以用二分加树状数组(时间复杂度log2n, 用平衡树也可以搞。。。）。
/*
* Problem: TJOI2013-day2-Sequence
* Author: Shun Yao
*/
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <assert.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <bitset>
#include <utility>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
//using namespace std;
const int MAXN = 100010;
int getInt() {
static char ch, f;
static int ret;
f = 1;
while (ch = getchar(), ch < '0' || ch > '9')
if (ch == '-')
f = 0;
ret = ch - '0';
while (ch = getchar(), ch >= '0' && ch <= '9')
ret = (ret << 1) + (ret << 3) + ch - '0';
return f ? ret : -ret;
}
int n;
int s[MAXN];
void modify(int x, int y) {
static int i;
for (i = x; i <= n; i += i & -i)
s[i] += y;
}
int query(int x) {
static int i, ret;
ret = 0;
for (i = x; i >= 1; i -= i & -i)
ret += s[i];
return ret;
}
struct SegTree {
int v;
SegTree *l, *r;
} *root;
void build(SegTree *&p, int l, int r) {
p = new SegTree();
p->v = 0;
p->l = p->r = NULL;
if (l == r)
return;
build(p->l, l, (l + r) >> 1);
build(p->r, ((l + r) >> 1) + 1, r);
}
void modify(SegTree *p, int l, int r, int x, int y) {
if (l == r) {
p->v = y;
return;
}
if (x <= (l + r) >> 1)
modify(p->l, l, (l + r) >> 1, x, y);
else
modify(p->r, ((l + r) >> 1) + 1, r, x, y);
p->v = std::max(p->l->v, p->r->v);
}
int query(SegTree *p, int l, int r, int x, int y) {
if (l == x && y == r)
return p->v;
if (y <= (l + r) >> 1)
return query(p->l, l, (l + r) >> 1, x, y);
else if (x > (l + r) >> 1)
return query(p->r, ((l + r) >> 1) + 1, r, x, y);
return std::max(query(p->l, l, (l + r) >> 1, x, (l + r) >> 1), query(p->r, ((l + r) >> 1) + 1, r, ((l + r) >> 1) + 1, y));
}
int main(/*int argc, char **argv*/) {
int i, j, mid, l, r, x[MAXN], a[MAXN], ans;
char v[MAXN];
freopen("sequence.in", "r", stdin);
freopen("sequence.out", "w", stdout);
n = getInt();
for (i = 1; i <= n; ++i)
x[i] = getInt();
memset(v, 0, sizeof v);
for (i = n; i >= 1; --i) {
l = 1;
r = n;
while (l <= r) {
mid = (l + r) >> 1;
j = mid - query(mid);
if (j == x[i] + 1) {
if (!v[mid]) {
v[a[i] = mid] = 1;
modify(mid, 1);
break;
} else
r = mid - 1;
continue;
}
if (j > x[i] + 1)
r = mid - 1;
else
l = mid + 1;
}
}
build(root, 1, n);
ans = 0;
for (i = 1; i <= n; ++i) {
j = query(root, 1, n, 1, a[i]);
modify(root, 1, n, a[i], j + 1);
if (ans < j + 1)
ans = j + 1;
printf("%d\n", ans);
}
fclose(stdin);
fclose(stdout);
return 0;
}
第二题：仔细分析可以发现应该按照a + b递增的顺序贪心出井，然后dp，f[i][j]代表前i个逃离了j个的剩余最大高度。
/*
* Problem: Dwarf
* Author: Shun Yao
*/
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <assert.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <bitset>
#include <utility>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
//using namespace std;
const int MAXN = 2222;
int n, f[MAXN];
struct Data {
int a, b;
} c[MAXN];
bool cmpa(Data a, Data b) {
return a.a + a.b < b.a + b.b;
}
int main(/*int argc, char **argv*/) {
int i, j, h;
freopen("dwarf.in", "r", stdin);
freopen("dwarf.out", "w", stdout);
scanf("%d", &n);
for (i = 1; i <= n; ++i)
scanf("%d%d", &c[i].a, &c[i].b);
scanf("%d", &h);
std::sort(c + 1, c + n + 1, cmpa);
for (i = 1; i <= n; ++i)
f[0] += c[i].a;
for (i = 1; i <= n; ++i)
f[i] = INT_MIN;
for (i = 1; i <= n; ++i)
for (j = i; j >= 1; --j)
if (f[j - 1] != INT_MIN && f[j - 1] + c[i].b >= h && f[j] < f[j - 1] - c[i].a)
f[j] = f[j - 1] - c[i].a;
for (i = n; i >= 0; --i)
if (f[i] >= 0)
break;
printf("%d", i);
fclose(stdin);
fclose(stdout);
return 0;
}
第三题：好吧，很明显二分图独立集（好像是好经典的题目啊！，据说匈牙利会被卡爆，我在BZOJ上交的。。。）
/*
* Problem: TJOI2013-day2-Attack
* Author: Shun Yao
*/
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <assert.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <bitset>
#include <utility>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
//using namespace std;
const int MAXN = 222;
const int dx[8] = {1, 1, -1, -1, 2, 2, -2, -2};
const int dy[8] = {2, -2, 2, -2, 1, -1, 1, -1};
int n, mat[MAXN * MAXN];
bool used[MAXN * MAXN];
class Edge {
public:
int v;
Edge *next;
Edge() {}
~Edge() {}
Edge(int V, Edge *ne) : v(V), next(ne) {}
} *g[MAXN * MAXN];
void add(int x, int y) {
g[x] = new Edge(y, g[x]);
}
bool find(int x) {
for (Edge *e = g[x]; e; e = e->next)
if (!used[e->v]) {
used[e->v] = 1;
if (!mat[e->v] || find(mat[e->v])) {
mat[e->v] = x;
return 1;
}
}
return 0;
}
int main(/*int argc, char **argv*/) {
int i, j, k, x, y, ans, sum;
char s[MAXN][MAXN];
freopen("attack.in", "r", stdin);
freopen("attack.out", "w", stdout);
scanf("%d", &n);
for (i = 1; i <= n; ++i)
scanf(" %s", s[i] + 1);
sum = 0;
for (i = 1; i <= n; ++i)
for (j = 1; j <= n; ++j) {
if (s[i][j] == '1')
continue;
++sum;
if ((i + j) & 1)
for (k = 0; k < 8; ++k) {
x = i + dx[k];
y = j + dy[k];
if (x < 1 || x > n || y < 1 || y > n || s[x][y] == '1')
continue;
add((i - 1) * n + j, (x - 1) * n + y);
}
}
ans = 0;
memset(mat, 0, sizeof mat);
for (i = 1; i <= n; ++i)
for (j = 1; j <= n; ++j)
if (s[i][j] == '0' && (i + j) & 1) {
memset(used, 0, sizeof used);
if (find((i - 1) * n + j))
++ans;
}
printf("%d\n", sum - ans);
fclose(stdin);
fclose(stdout);
return 0;
}