标签：style io strong for ar amp sp on size
A.ZOJ 3666 Alice and Bob
组合博弈，SG函数应用#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 10000 + 100;
int SG[maxn];
vector<int> g[maxn];
int mex(int u) { //minimal excludant
if(SG[u]!=-1) return SG[u];
int i;
bool vis[maxn];
memset(vis, 0, sizeof vis );
for(i=0; i<g[u].size(); ++i) {
vis[mex(g[u][i])] = true;
}
for(i=0; vis[i]; ++i);
return SG[u] = i;
}
int main() {
int n, c, x, q, m;
int cas= 1;
while(~scanf("%d", &n)) {
for(int i=0; i<=n; ++i) g[i].clear();
for(int i=1; i<n; ++i) {
scanf("%d", &c);
while(c--) {
scanf("%d", &x);
g[i].push_back(x);
}
}
memset(SG, -1, sizeof SG );
printf("Case %d:\n", cas++);
scanf("%d", &q);
while(q--) {
scanf("%d", &m);
int ans = 0;
while(m--) {
scanf("%d", &x);
ans ^= mex(x);
}
if(ans) puts("Alice");
else puts("Bob");
}
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1010;
const int maxm = 40000 + 100;
const int INF = 1e9;
struct Edge{
int to, next;
int w;
}edge[maxm];
int head[maxn], tot;
void init()
{
tot = 0;
memset(head, -1, sizeof head );
}
void addedge(int u, int v, int w){
edge[tot].to = v;
edge[tot].next = head[u];
edge[tot].w = w;
head[u] = tot++;
}
bool vis[maxn];
int cnt[maxn];
int dist[maxn];
bool spfa(int n)
{
memset(vis, false, sizeof vis );
for(int i=0; i<=n; ++i) dist[i] = INF;
memset(cnt, 0, sizeof cnt );
queue<int> q;
q.push(0);
vis[0] = true;
cnt[0] = 1;
dist[0] = 0;
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = false;
for(int i=head[u]; i!=-1; i=edge[i].next){
int v = edge[i].to;
int w = edge[i].w;
if(dist[v] > dist[u] + w){
dist[v] = dist[u] + w;
if(!vis[v]){
vis[v] = true;
q.push(v);
if(++cnt[v] > n+1) return false;
}
}
}
}
return true;
}
int main()
{
int n, m;
while(~scanf("%d%d", &n, &m)){
init();
for(int i=1; i<=n; ++i){
addedge(i-1, i, 10000);
addedge(i, i-1, 10000);
}
int l, r, a, b;
while(m--){
scanf("%d%d%d%d", &l, &r, &a, &b);
addedge(l-1, r, b);
addedge(r, l-1, -a);
}
if(!spfa(n)){
printf("The spacecraft is broken!\n");
continue;
}
for(int i=1; i<=n; ++i){
printf("%d", dist[i]-dist[i-1]);
if(i<n) printf(" ");
else printf("\n");
}
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
const int maxn = 9;
int z[maxn+10], s[maxn+10], p[maxn+10], m[maxn+10];
char str[100];
bool seven()
{
int cnt = 0;
for(int i=1; i<=9; ++i)
{
if(z[i]==2) cnt++;
if(s[i]==2) cnt++;
if(p[i]==2) cnt++;
if(m[i]==2) cnt++;
}
return cnt==7;
}
bool thirteen()
{
int cnt = m[1] + m[9] + p[1] + p[9] + s[1] + s[9] + z[1] + z[2] + z[3] + z[4] + z[5]
+ z[6] + z[7];
if(m[1]>=1&&m[9]>=1&&p[1]>=1&&p[9]>=1&&s[1]>=1&&s[9]>=1&&z[1]>=1&&z[2]>=1&&z[3]>=1&&z[4]>=1
&&z[5]>=1&&z[6]>=1&&z[7]>=1){
return cnt == 14;
}
return 0;
}
int main()
{
while(~scanf("%s", str)) {
int n = strlen(str);
memset(z, 0, sizeof z );
memset(s, 0, sizeof s );
memset(p, 0, sizeof p );
memset(m, 0, sizeof m );
for(int i=0; i<n; i+=2) {
int num = str[i] - '0';
if(str[i+1]=='z')
z[num]++;
else if(str[i+1]=='s')
s[num]++;
else if(str[i+1]=='m')
m[num]++;
else if(str[i+1]=='p')
p[num]++;
}
if(seven()) {
printf("Seven Pairs\n");
} else if(thirteen()) {
printf("Thirteen Orphans\n");
} else printf("Neither!\n");
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int main () {
int n;
while (~scanf("%d",&n)) {
LL maxx = 0, sum = 0;
for (int i=1; i<=n; i++) {
LL a,b;
scanf("%I64d%I64d",&a, &b);
maxx=max(maxx,a-b);
sum+=a-b;
}
cout<<sum<<" "<<maxx<<endl;
if ((sum%2==1) ||maxx*2>sum) printf("NO\n");
else printf("YES\n");
}
return 0;
}
H.ZOJ 3673 1729
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100 + 10;
set<string> S[maxn];
map<string,int> M;
char buf[1001], str[1001];
vector<int> vc;
int n, m;
int main() {
while(~scanf("%d", &n)) {
M.clear();
string ss;
for(int i=0; i<n; ++i) {
S[i].clear();
cin>>ss;
M[ss] = i;
getchar();
gets(buf);
char *p = strtok(buf," ");
while(p) {
sscanf(p, "%s", str );
S[i].insert(string(str));
p = strtok(NULL, " ");
}
}
scanf("%d", &m);
getchar();
while(m--) {
vc.clear();
gets(buf);
char *p = strtok(buf, " ");
while(p) {
sscanf(p, "%s", str );
vc.push_back( M[string(str)] );
p = strtok(NULL, " ");
}
vector<string> ans;
for(set<string>::iterator it = S[vc[0]].begin(); it!=S[vc[0]].end(); ++it) {
string tmp = *it;
bool found = true;
for(int i=1; i<vc.size(); ++i) {
if(S[vc[i]].find(tmp) == S[vc[i]].end()) {
found = false;
break;
}
}
if(found) ans.push_back(tmp);
}
sort(ans.begin(), ans.end());
if(ans.size()==0) puts("NO");
else {
for(int i=0; i<ans.size(); ++i) {
cout<<ans[i];
if(i<ans.size()-1) cout<<" ";
else cout<<endl;
}
}
}
}
return 0;
}
状态压缩DP: 状态:dp[1<<M-1]，结果是dp[0]，状态转移即 枚举指甲剪剪的位置和顺序，然后产生新状态。
#include<bits/stdc++.h>
using namespace std;
const int INF = 1e9;
int dp[1<<21];
int main() {
int n, m;
char s[20];
while(~scanf("%d", &n)) {
scanf("%s", s);
scanf("%d", &m);
int clr = 0, rclr = 0;
for(int i=0; i<n; ++i) {
clr = (clr<<1) + (s[i]=='*');
rclr = (rclr<<1) + (s[n-i-1]=='*');
}
for(int i=0; i<=(1<<m); ++i) dp[i] = INF;
dp[(1<<m)-1] = 0;
for(int i=(1<<m)-1; i>=0; --i)
if(dp[i]!=INF) {
for(int j=0; j<m; ++j) {
dp[i& (~(clr<<j))] = min(dp[i&(~(clr<<j))], dp[i] + 1);
dp[i& (~(rclr<<j))] = min(dp[i&(~(rclr<<j))], dp[i]+1);
dp[i& (~(clr>>j))] = min(dp[i&(~(clr>>j))], dp[i] + 1);
dp[i& (~(rclr>>j))] = min(dp[i&(~(rclr>>j))], dp[i]+1);
}
}
if(dp[0]!=INF) {
printf("%d\n", dp[0]);
} else printf("-1\n");
}
return 0;
}
标签：style io strong for ar amp sp on size
原文地址：http://blog.csdn.net/yew1eb/article/details/38865593
