爆零记
- 本来都忘了有这个比赛emmm
然后下午其实也没时间写的样子，
可是一直发短信提醒，，
所以最后还是抽了小半个小时敲了两份模板
- 但是，，大概是我菜把，
两份都只能过样例。。。。
而且也调试不出个所以然来
包括拿到题解都觉得我思路没啥问题。。
T1、五子棋
- 题意：给出一个五子棋盘，判断下一步能不能赢，下哪里能赢。
- 根据棋子数量可以得知下一步谁下棋。然后枚举每个位置判断能否获胜即可。复杂度O(25*25*5)感觉挺稳的，可是不知道为啥只过了样例
#include
#include
using namespace std;
char c[50][50];
int main(){
int numo = 0, numx = 0;
for(int i = 1; i <= 25; i++){
for(int j = 1; j <= 25; j++){
c[i][j] = getchar();
if(c[i][j]=='o') numo++;
if(c[i][j]=='x') numx++;
}
getchar();
}
#include
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << "Line(" << __LINE__ << ") -> " << #x << " is " << x << endl
using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;
const int INF = 1e9 + 7;
char a[25][25], who;
bool yes() {
for(int i = 0; i <= 20; i++) {
for(int j = 0; j < 25; j++) {
bool ok = true;
for(int k = 0; k < 5; k++)
if(a[i+k][j]!=who) ok = false;
if(ok) return true;
}
}
for(int i = 0; i < 25; i++) {
for(int j = 0; j <= 20; j++) {
bool ok = true;
for(int k = 0; k < 5; k++)
if(a[i][j+k]!=who) ok = false;
if(ok) return true;
}
}
for(int i = 0; i <= 20; i++) {
for(int j = 0; j <= 20; j++) {
bool ok = true;
for(int k = 0; k < 5; k++)
if(a[i+k][j+k]!=who) ok = false;
if(ok) return true;
}
}
for(int i = 4; i < 25; i++) {
for(int j = 0; j <= 20; j++) {
bool ok = true;
for(int k = 0; k < 5; k++)
if(a[i-k][j+k]!=who) ok = false;
if(ok) return true;
}
}
return false;
}
signed main()
{
IO_OP;
int cnt = 0;
for(int i = 0; i < 25; i++)
for(int j = 0; j < 25; j++) {
cin >> a[i][j];
if(a[i][j] != '.') cnt++;
}
who = 'x';
if(cnt&1) who = 'o';
bool y = 0;
for(int i = 0; i < 25; i++) {
for(int j = 0; j < 25; j++) {
if(a[i][j] == '.') {
a[i][j] = who;
if(yes()) {
cout << i << " " << j << endl;
y = 1;
}
a[i][j] = '.';
}
}
}
if(!y) cout << "tie" << endl;
}
T2、染色（简单）
- 题意：给出一个长为n的序列，以及m段区间。对序列的每个元素染色成黑或白分别可以获得bi和wi的美感，若某区间染成相同则可额外获得ci的美感，求美感最大。
- 考虑没有区间的情况，直接扫一遍统计bi和wi单个元素最大即可。而加上区间则意味着有可能获得更高的美感，即当区间全染白+ci>区间染最大的时候，用后者更新即可。用前缀和维护可以O(1)求区间和，直接统计即可。整体复杂度是O(n)的，常数比较大，应该能过。
#include
#include
using namespace std;
const int maxn = 300050;
typedef long long LL;
int n, m, b[maxn], w[maxn], x[maxn];
LL sb[maxn], sw[maxn], sx[maxn];
int main(){
cin>>n>>m;
for(int i = 1; i <= n; i++){
cin>>b[i];
sb[i] = sb[i-1]+b[i];
}
for(int i = 1; i <= n; i++){
cin>>w[i];
x[i] = max(b[i],w[i]);
sw[i] = sw[i-1]+w[i];
sx[i] = sx[i-1]+x[i];
}
int ans = sx[n];
for(int i = 1; i <= m; i++){
int t, l, r, c;
cin>>t>>l>>r>>c;
if(t==1){
int tmp = sb[r]-sb[l-1];
if(tmp+c>sx[r]-sx[l-1]){
ans = ans-(sx[r]-sx[l-1])+tmp+c;
}
}
if(t==2){
int tmp = sw[r]-sw[l-1];
if(tmp+c>sx[r]-sx[l-1]){
ans = ans-(sx[r]-sx[l-1])+tmp+c;
}
}
}
cout<<ans;
return 0;
}
#include
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << "Line(" << __LINE__ << ") -> " << #x << " is " << x << endl
#define int ll
using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;
const int INF = 1e9 + 7, N = 3e5 + 7;
int b[N], w[N], bsum[N], wsum[N], mx[N], sum[N], yes[N];
signed main()
{
IO_OP;
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; i++) {
cin >> b[i];
bsum[i] = bsum[i - 1] + b[i];
}
for(int i = 1; i <= n; i++) {
cin >> w[i];
wsum[i] = wsum[i - 1] + w[i];
}
for(int i = 1; i <= n; i++) {
mx[i] = max(b[i], w[i]);
sum[i] = mx[i] + sum[i - 1];
}
int ans = 0;
for(int i = 1; i <= m; i++) {
int t, l, r, c;
cin >> t >> l >> r >> c;
for(int j = l; j <= r; j++) yes[j] = 1;
int cur = sum[r] - sum[l-1];
if(t == 1) {
cur = max(cur, bsum[r] - bsum[l-1] + c);
} else {
cur = max(cur, wsum[r] - wsum[l-1] + c);
}
ans += cur;
}
for(int i = 1; i <= n; i++) if(!yes[i]) ans += mx[i];
cout << ans << endl;
}
#include
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << "Line(" << __LINE__ << ") -> " << #x << " is " << x << endl
#define int ll
using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;
const int INF = 1e9 + 7, N = 3e5 + 7;
int b[N], w[N], bsum[N], wsum[N], mx[N], sum[N], yes[N];
signed main()
{
IO_OP;
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; i++) {
cin >> b[i];
bsum[i] = bsum[i - 1] + b[i];
}
for(int i = 1; i <= n; i++) {
cin >> w[i];
wsum[i] = wsum[i - 1] + w[i];
}
for(int i = 1; i <= n; i++) {
mx[i] = max(b[i], w[i]);
sum[i] = mx[i] + sum[i - 1];
}
int ans = 0;
for(int i = 1; i <= m; i++) {
int t, l, r, c;
cin >> t >> l >> r >> c;
for(int j = l; j <= r; j++) yes[j] = 1;
int cur = sum[r] - sum[l-1];
if(t == 1) {
cur = max(cur, bsum[r] - bsum[l-1] + c);
} else {
cur = max(cur, wsum[r] - wsum[l-1] + c);
}
ans += cur;
}
for(int i = 1; i <= n; i++) if(!yes[i]) ans += mx[i];
cout << ans << endl;
}
T3、染色（中等）
#include
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << "Line(" << __LINE__ << ") -> " << #x << " is " << x << endl
#define int ll
using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;
const int INF = 1e9 + 7, N = 3e5 + 7;
int b[N], w[N], bsum[N], wsum[N], mx[N], sum[N], dp[N];
V<pi> bb[N], ww[N];
int bmore[N], wmore[N];
signed main()
{
IO_OP;
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; i++) {
cin >> b[i];
bsum[i] = bsum[i - 1] + b[i];
}
for(int i = 1; i <= n; i++) {
cin >> w[i];
wsum[i] = wsum[i - 1] + w[i];
}
for(int i = 1; i <= n; i++) {
mx[i] = max(b[i], w[i]);
sum[i] = mx[i] + sum[i - 1];
}
vi v;
v.PB(1), v.PB(n+1);
for(int i = 0; i < m; i++) {
int t, l, r, c;
cin >> t >> l >> r >> c;
if(t == 1) {
bb[r].EB(l, c);
} else {
ww[r].EB(l, c);
}
v.PB(l);
v.PB(r+1);
}
sort(ALL(v));
v.resize(unique(ALL(v))-v.begin());
int nn = v.size() - 1;
auto getmx = [&](int l, int r) {
int lb = v[l], rb = v[r + 1] - 1;
return sum[rb] - sum[lb-1];
};
auto getb = [&](int l, int r) {
int lb = v[l], rb = v[r + 1] - 1;
return bsum[rb] - bsum[lb-1];
};
auto getw = [&](int l, int r) {
int lb = v[l], rb = v[r + 1] - 1;
return wsum[rb] - wsum[lb-1];
};
for(int i = 0; i < nn; i++) {
int r = v[i + 1] - 1;
for(auto p:bb[r]) {
bmore[p.F] += p.S;
}
for(auto p:ww[r]) {
wmore[p.F] += p.S;
}
dp[i] = (i-1 < 0 ? 0 : dp[i-1]) + getmx(i, i);
int more = 0;
for(int j = i; j >= 0; j--) {
more += bmore[v[j]];
dp[i] = max(dp[i], (j-1 < 0 ? 0 : dp[j-1]) + getb(j, i) + more);
}
more = 0;
for(int j = i; j >= 0; j--) {
more += wmore[v[j]];
dp[i] = max(dp[i], (j-1 < 0 ? 0 : dp[j-1]) + getw(j, i) + more);
}
}
cout << dp[nn-1] << endl;
}
