2017"百度之星"程序设计大赛 - 资格赛
度度熊保护村庄
Accepts: 47
Submissions: 900
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
Problem Description
哗啦啦村袭击了喵哈哈村！ 度度熊为了拯救喵哈哈村，带着自己的伙伴去救援喵哈哈村去了！度度熊与伙伴们很快的就过来占据了喵哈哈村的各个军事要地，牢牢的守住了喵哈哈村。 但是度度熊发现，这是一场旷日持久的战斗，所以度度熊决定要以逸待劳，保存尽量多的体力，去迎战哗啦啦村的战士。 于是度度熊决定派尽量多的人去休息，但是同时也不能松懈对喵哈哈村的保护。 换句话而言，度度熊希望尽量多的人休息，而且存在一个包围圈由剩下的人组成，且能够恰好的包围住喵哈哈村的所有住房（包括边界）。 请问最多能让多少个人休息呢？
Input
本题包含若干组测试数据。 第一行一个整数n，表示喵哈哈村的住房数量。 接下来n行，每行两个整数(x1[i],y1[i])，表示喵哈哈村的住房坐标。 第n+1行一个整数m，表示度度熊的士兵数量。 接下来m行，每行两个整数(x2[i],y2[i])，表示度度熊伙伴的坐标。 满足: 1<=n,m<=500 -10000<=x1[i],x2[i],y1[i],y2[i]<=10000
Output
请输出最多的人员休息的数目。 如果无法保护整个村庄的话，输出"ToT"
Sample Input
Copy
2 1 1 2 2 4 0 0 0 4 4 2 4 0 1 1 1 2 0 0 0 1
Sample Output
Copy
1 ToT
#include
#include
#define FOR(i, l, r) for(int i = l; i <= r; i++)
using namespace std;
const int N = 510;
struct edge{int to, next;} e[N * N];
int n, m, ans, fl, cnt;
int vis[N], head[N], h[N];
struct poi{int x, y;} a[N], b[N];
poi operator - (poi a, poi b) {poi c; c.x = a.x - b.x; c.y = a.y - b.y; return c;}
int operator * (poi a, poi b) {return a.x * b.y - a.y * b.x;}
void ins(int x, int y)
{
e[++cnt].to = y; e[cnt].next = head[x]; head[x] = cnt;
}
int bfs(int y)
{
FOR(i, 1, m) vis[i] = 0;
int l = 0, r = 1; vis[y] = 1; h[1] = y;
while (l < r)
{
int x = h[++l];
for(int i = head[x]; i; i = e[i].next)
{
if (e[i].to == y) return vis[x];
if (!vis[e[i].to])
{
vis[e[i].to] = vis[x] + 1;
h[++r] = e[i].to;
}
}
}
return m + 1;
}
int main()
{
while (scanf("%d", &n) != EOF)
{
FOR(i, 1, n) scanf("%d%d", &a[i].x, &a[i].y);
scanf("%d", &m);
FOR(i, 1, m) scanf("%d%d", &b[i].x, &b[i].y);
cnt = 0; FOR(i, 1, m) head[i] = 0;
FOR(i, 1, m) FOR(j, i + 1, m)
{
fl = 1;
FOR(k, 1, n) if ((b[j] - b[i]) * (a[k] - b[i]) < 0) {fl = 0; break;}
if (fl) ins(i, j);
fl = 1;
FOR(k, 1, n) if ((b[i] - b[j]) * (a[k] - b[j]) < 0) {fl = 0; break;}
if (fl) ins(j, i);
}
ans = m + 1;
FOR(i, 1, m) ans = min(ans, bfs(i));
if (ans > m) puts("ToT"); else printf("%d\n", m - ans);
}
return 0;
}
度度熊的王国战略
Accepts: 674
Submissions: 6056
Time Limit: 40000/20000 MS (Java/Others)
Memory Limit: 32768/132768 K (Java/Others)
Problem Description
度度熊国王率领着喵哈哈族的勇士，准备进攻哗啦啦族。 哗啦啦族是一个强悍的民族，里面有充满智慧的谋士，拥有无穷力量的战士。 所以这一场战争，将会十分艰难。 为了更好的进攻哗啦啦族，度度熊决定首先应该从内部瓦解哗啦啦族。 第一步就是应该使得哗啦啦族内部不能同心齐力，需要内部有间隙。 哗啦啦族一共有n个将领，他们一共有m个强关系，摧毁每一个强关系都需要一定的代价。 现在度度熊命令你需要摧毁一些强关系，使得内部的将领，不能通过这些强关系，连成一个完整的连通块，以保证战争的顺利进行。 请问最少应该付出多少的代价。
Input
本题包含若干组测试数据。 第一行两个整数n，m，表示有n个将领，m个关系。 接下来m行，每行三个整数u,v,w。表示u将领和v将领之间存在一个强关系，摧毁这个强关系需要代价w 数据范围： 2<=n<=3000 1<=m<=100000 1<=u,v<=n 1<=w<=1000
Output
对于每组测试数据，输出最小需要的代价。
Sample Input
Copy
2 1 1 2 1 3 3 1 2 5 1 2 4 2 3 3
Sample Output
Copy
1 3
#include
#include
#include
#include
#include
#include
using namespace std;
vector > g[3001];
int main()
{
srand((unsigned)time(NULL));
int n,m,u,v,w,i,j,ans,anst,times,cost[3001],lsize,rsize;
bool tag[3001],rep;
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=n;++i)
{
g[i].clear();
}
while(m--)
{
scanf("%d%d%d",&u,&v,&w);
if(u==v)
{
continue;
}
g[u].push_back(make_pair(v,w));
g[v].push_back(make_pair(u,w));
}
ans=1e8;
times=0;
do
{
++times;
memset(tag+1,false,sizeof(bool)*n);
tag[times]=true;
lsize=1;
rsize=n-1;
anst=0;
memset(cost+1,0,sizeof(int)*n);
for(i=1;i<=n;++i)
{
for(j=0;j>=1;
for(i=1;i<=n;++i)
{
cost[i]>>=1;
}
rep=true;
while(rep)
{
rep=false;
for(i=2;i<=n;++i)
{
if(cost[i]<0 && ((!tag[i]) || lsize>1) && ((tag[i]) || rsize>1))
{
rep=true;
anst+=cost[i];
if(tag[i])
{
--lsize;
++rsize;
}
else
{
++lsize;
--rsize;
}
for(j=0;j 度度熊与邪恶大魔王 Accepts: 3666
Submissions: 22474
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
Problem Description
度度熊为了拯救可爱的公主，于是与邪恶大魔王战斗起来。
邪恶大魔王的麾下有n个怪兽，每个怪兽有a[i]的生命值，以及b[i]的防御力。
度度熊一共拥有m种攻击方式，第i种攻击方式，需要消耗k[i]的晶石，造成p[i]点伤害。
当然，如果度度熊使用第i个技能打在第j个怪兽上面的话，会使得第j个怪兽的生命值减少p[i]-b[j]，当然如果伤害小于防御，那么攻击就不会奏效。
如果怪兽的生命值降为0或以下，那么怪兽就会被消灭。
当然每个技能都可以使用无限次。
请问度度熊最少携带多少晶石，就可以消灭所有的怪兽。
Input
本题包含若干组测试数据。
第一行两个整数n，m,表示有n个怪兽，m种技能。
接下来n行，每行两个整数，a[i],b[i]，分别表示怪兽的生命值和防御力。
再接下来m行，每行两个整数k[i]和p[i]，分别表示技能的消耗晶石数目和技能的伤害值。
数据范围:
1<=n<=100000
1<=m<=1000
1<=a[i]<=1000
0<=b[i]<=10
0<=k[i]<=100000
0<=p[i]<=1000
Output
对于每组测试数据，输出最小的晶石消耗数量，如果不能击败所有的怪兽，输出-1
Sample Input
1 2
3 5
7 10
6 8
1 2
3 5
10 7
8 6
Sample Output
6
18
#include
#include
#include
#include
using namespace std;
typedef long long ll;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline bool read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1; else if (c==EOF) return 0;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b; return 1;
}
const int N=100005;
int n,m;
ll f[11][1005];
int k[N],p[N];
inline void DP(int b,ll *f){
for (int i=1;i<=1000;i++) f[i]=1LL<<50;
f[0]=0;
for (int i=1;i<=m;i++)
if (p[i]>b)
for (int j=0;j<=1000;j++)
f[min(j+p[i]-b,1000)]=min(f[min(j+p[i]-b,1000)],f[j]+k[i]);
for (int i=1000-1;~i;i--)
f[i]=min(f[i+1],f[i]);
}
int a[N],b[N];
int main(){
while (read(n) && read(m)){
for (int i=1;i<=n;i++) read(a[i]),read(b[i]);
for (int i=1;i<=m;i++) read(k[i]),read(p[i]);
for (int i=0;i<=10;i++)
DP(i,f[i]);
ll Ans=0;
for (int i=1;i<=n;i++)
Ans+=f[b[i]][a[i]];
if (Ans>=1LL<<50)
printf("-1\n");
else
printf("%I64d\n",Ans);
}
return 0;
}
度度熊的午饭时光
Accepts: 823
Submissions: 9241
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
Problem Description
度度熊最期待每天的午饭时光，因为早饭菜品清淡，晚饭减肥不敢吃太多（胖纸的忧伤T.T）。
百度食堂的午餐超级丰富，祖国各大菜系应有尽有，度度熊在每个窗口都有爱吃的菜品，而且他还为喜爱的菜品打了分，吃货的情怀呀(>.<)。
但是，好吃的饭菜总是很贵，每天的午饭预算有限，请帮度度熊算一算，怎样打饭才能买到的最好吃的饭菜?(不超过预算、不重样、午餐等分最高的情况下，选择菜品序号加和最小，加和相等时字典序最小的组合)
Input
第一行一个整数T，表示T组数据。每组测试数据将以如下格式从标准输入读入：
B
N
score_1 cost_1
score_2 cost_2
:
score_N cost_N
第一行，正整数B（0 <= B <= 1000），代表午餐的预算。
第二行，正整数N (0 <= N <= 100)，代表午餐可选的菜品数量
从第三行到第 (N + 2) 行，每行两个正整数，以空格分隔，score_i表示菜品的得分，cost_i表示菜品的价格(0 <= score_i, cost_i <= 100)。
Output
对于每组数据，输出两行：第一行输出："Case #i:"。i代表第i组测试数据。第二行输出菜品的总得分和总花费，以空格分隔。第三行输出所选菜品的序号，菜品序号从1开始，以空格分隔。
Sample Input
2 29 6 9 10 3 4 6 5 7 20 10 9 15 11 0 2 2 23 10 12
Sample Output
Case #1: 34 29 2 3 5 6 Case #2: 0 0
#include
#include
#include
#include
#include
#include
#include
#include 寻找母串 Accepts: 211
Submissions: 2019
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
Problem Description
对于一个串S，当它同时满足如下条件时，它就是一个01偏串： 1、只由0和1两种符组成； 2、在S的每一个前缀中，0的个数不超过1的个数； 3、S中0的个数和1的个数相等。
现在给定01偏串S，请计算一下S在所有长度为n的01偏串中作为子串出现的次数的总和。 由于结果比较大，结果对1e9+7取余后输出。
样例解释： 在第二个样例中，长度为4的偏串共两个1010，1100。10在1010中出现了两次，在1100中出现了1次。所以答案是3。
Input
第一行给出一个整数T(1<=T<=40),表示测试数据的数目。 每一组测试包含一个整数n和字符串S，中间用空格分开。(1<=|S|<=100000，1<=n<=1000000000)
输入保证S是一个01偏串。
Output
对于每一组数据，输出一个整数占一行，表示答案。
Sample Input
2
2 10
4 10
Sample Output
1
3#include
#include
const int K=500000,P=1000000007;
int Case,n,m;char a[K];
int f[]={1,154425679,641102369,448516387,578095319,119948046,5832229,929795683,259081142,59433257,974067448,261465122,316220877,663134272,690120224,304661605,251368199,704262886,980250487,386267146,682498929,681184885,134623568,427907847,95936601,140928835,933097914,391177708,167332441,739650465,598816162,292015879,336060741,810640932,248744620,892704918,626497524,118508240,288843364,406755567,491101308,668723114,245341950,560364972,565768255,913230882,246899319,756627904,968999,7998896,586350670,328019447,638587686,497857012,881746146,526658674,19426633,560607302,850500036,916364742,76479948,732041432,268124147,30473806,842267748,878381584,886294336,73995168,485348706,771153434,463847391,72518508,544075857,264638935,898187927,468116664,798967520,359622660,82926604,59761470,723816384,320608814,156530778,278224088,721996174,222176885,299085602,814440466,323604647,359932651,172827403,213590923,398699886,983975503,530389102,619965361,294587621,221896631,813805606,676243000,67347853,557470122,497478507,494243939,196447201,726686688,722054885,435396042,228338256,574429123,407719831,583676448,762479457,342545697,746536789,3669475,811667359,249780014,778773518,306921081,27368307,467386519,438371670,44635470,59469516,498440078,5974669,9221150,766196482,679975009,606322308,147307265,86609485,435993596,889750731,607649469,340941507,886844916,371263376,198057134,625544428,527225600,788878910,743572663,808412394,603835511,996952918,409420559,585237443,307534126,1669644,210601394,361786913,470618701,480748381,550969600,595143852,901560583,837229828,727223125,199888908,180269718,526807168,899715096,579691190,775147177,145404005,401312408,459188207,552930818,534491822,669251006,439729802,504703143,840398449,624903360,899297830,556957390,235861787,850796519,888050723,834568419,656116726,196710441,736550105,857549983,440902696,63145422,85990869,807209645,884343068,132114154,56305184,973762120,973478770,270603374,168891766,217608725,804805577,641317239,927880474,157803266,876297919,189324843,934814019,374934947,676405347,30657162,567277637,686637388,112249297,631678522,44930135,954976637,39417871,977168136,47401357,809639670,108819476,250361638,281863274,410799320,60168088,142840220,692636218,261319829,432775082,200805416,14235602,616435469,770511792,174624465,400295761,580310773,697066277,102309489,421835306,228266417,220108638,334251014,661224977,812005526,261799937,356438399,168203998,835330203,802214249,316146626,544064410,877125482,935080803,576478595,583967898,885117574,211768084,672118586,751231582,192990501,972424306,125666214,623534362,362493992,335160196,924982465,243276029,85322142,554749550,437447268,60050552,733942454,797848181,643938537,395891998,435508211,172428290,422713438,159554990,812391270,887420150,449205554,970055531,222343461,250388809,689850274,487998999,748359168,856259313,29781141,82104855,543656809,232253360,132581614,513365505,443711760,244109365,956163585,1559745,193109503,695345956,36042990,261384175,953585846,849009131,600281990,323214113,762863177,747664143,388430024,444090941,396611787,659224434,991836842,80729842,231753353,570033864,24300293,664989237,653829979,827348878,7704403,195888993,30813167,576798521,947304403,457882808,137550654,731551699,995246811,212938473,593989303,509096183,478919708,827544702,602197504,678320208,496537397,677711203,386677716,289752035,456967785,66404266,208554578,555972231,680796549,195290384,364911332,97136305,294375015,349551356,290196031,785113347,297880214,83489485,644670437,66247239,135500451,52167191,498572695,307390891,499606972,547665832,932668364,143066173,499761057,350016754,113905458,917404120,493905303,296269301,86726457,996122673,809339058,23015220,835803486,602139210,670821828,748566338,631823487,187348575,382748764,109838563,94904185,574053420,973865472,105574531,752194712,304173654,83284607,542432219,536973025,34538816,283727326,325636655,243291133,437843114,312536013,630621321,682680116,26853683,611209607,933245637,894820080,616368450,198631047,238971581,595959409,511371690,683585700,557301633,253427558,911398531,246476028,848952161,272634280,958992544,325284740,925152039,495860053,914456118,841108662,724691727,445713321,636817583,814845293,238087006,170238260,946237212,457433989,910291942,599284582,114985663,17739814,492237273,194710256,450387329,471822076,834860913,831422618,763017204,100582220,368925948,128260642,475812562,917840849,740594930,51595198,45060610,994271859,806047532,476604857,464456846,835245756,172115341,545338291,75307702,344053349,116261993,945918607,562519302,281653965,268838846,635794655,173784895,258300540,243624360,722807834,61570384,691306302,481661251,752609613,938269070,381369130,95182730,885113729,91068149,758827428,115435332,617737077,495022305,641462345,136026497,537165746,506496856,997042337,710729672,185163716,113570024,90871114,366384665,783454081,564758715,75276833,270239666,203381481,277118392,639320111,79874094,366910724,702807165,582981397,112390913,245293699,730341625,338718815,103056890,1840485,677948390,391004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int po(int a,int b){int t=1;for(;b;b>>=1,a=1LL*a*a%P)if(b&1)t=1LL*t*a%P;return t;}
int fac(int n){
int t=n/K*K,ret=f[n/K];
while(n>t){
ret=1LL*ret*n%P;
n--;
}
return ret;
}
int cal(int n){
int t=fac(n);
return 1LL*fac(n+n)*po(t,P-2)%P*po(t,P-2)%P*po(n+1,P-2)%P;
}
int solve(){
if(n