【11/11】模拟赛
官方的链接：http://www.byvoid.com/blog/byvoid-wow-stage-3/
本博客不再给出题目。
第一题 彩色穿孔卡片
用的O(N^2)的算法。类似USACO的浮水法。
第二题 艾萨拉的激流
f[i][j] = max｛
f[i – 1][j – 1]
f[i – 1][j]
f[i – 1][j + 1]
} + a[i][j]
第三题 阿鲁高的阴谋
设f[i][j]是第i天过去后，存下了j的燃料。设l是第i天骗的燃料数。k是上一天剩下的燃料数。
k + l = n[i] + j，推出k = n[i] + j – l。
f[i][j] = min{f[i][j],f[i – 1][k] + l * b[i] + j * w}。
第四题 潜入辛迪加
多状态的宽搜。用state表示所有钥匙的状态。第i个钥匙对应的位置为1 << (i – 1)。
v[state][i][j]代表state状态下，(i,j)这个位置是否到过。
代码
第一题
#include <stdio.h>
#include <iostream>
#define MAXN 200001
using namespace std;
struct ss {
int l,r;
} a[MAXN];
bool have[MAXN];
int ans,n;
void flo(int l,int r,int x,int co) {
if (l >= r)
return;
if (have[co])
return;
if (x > n) {
have[co] = 1;
++ans;
return;
}
if (have[co])
return;
if (a[x].l >= l)
flo(l,min(a[x].l,r),x + 1,co);
if (a[x].r <= r)
flo(max(a[x].r,l),r,x + 1,co);
}
int main() {
freopen("punch.in","r",stdin);
freopen("punch.out","w",stdout);
scanf("%d",&n);
for (int i = 1;i <= n;++i)
scanf("%d%d",&a[i].l,&a[i].r);
for (int i = 1;i <= n;++i)
flo(a[i].l,a[i].r,i + 1,i);
printf("%d\n",ans);
return 0;
}
第二题
#include <stdio.h>
#include <string.h>
#define MAXN 1010
int f[2][MAXN],a[MAXN];
int n,m,ans,now,last;
int main() {
freopen("azshara.in","r",stdin);
freopen("azshara.out","w",stdout);
scanf("%d%d",&m,&n);
now = 1;
for (int i = 1;i <= n;++i) {
last = now;
now ^= 1;
for (int j = 1;j <= m;++j)
scanf("%d",&a[j]);
memset(f[now],255,sizeof(f[now]));
for (int j = 1;j <= m;++j)
if (a[j] != -1) {
if ((j - 1 > 0) && (f[last][j - 1] != -1) && (f[last][j - 1] + a[j] > f[now][j]))
f[now][j] = f[last][j - 1] + a[j];
if ((f[last][j] != -1) && (f[last][j] + a[j] > f[now][j]))
f[now][j] = f[last][j] + a[j];
if ((j + 1 <= m) && (f[last][j + 1] != -1) && (f[last][j + 1] + a[j] > f[now][j]))
f[now][j] = f[last][j + 1] + a[j];
}
}
for (int j = 1;j <= m;++j)
if (f[now][j] > ans)
ans = f[now][j];
printf("%d\n",ans);
return 0;
}
第三题
#include <stdio.h>
#define MAXT 1010
#define MAXINT 10000000
int b[MAXT],n[MAXT];
int t,v,a,w,x,y;
int f[MAXT][110],d[MAXT][110],ans[MAXT];
int main() {
freopen("arugal.in","r",stdin);
freopen("arugal.out","w",stdout);
scanf("%d%d%d%d",&t,&v,&a,&w);
for (int i = 0;i <= t;++i)
for (int j = 0;j <= v;++j)
f[i][j] = MAXINT;
f[0][0] = 0;
for (int i = 1;i <= t;++i) {
scanf("%d%d",&n[i],&b[i]);
for (int j = 0;j <= v;++j)
for (int l = 0;l <= a;++l) {
int k = n[i] + j - l;
if ((k >= 0) && (k <= v))
if (f[i - 1][k] + j * w + b[i] * l < f[i][j]) {
f[i][j] = f[i - 1][k] + j * w + b[i] * l;
d[i][j] = l;
}
}
}
int mi = MAXINT;
y = 0;
mi = f[t][0];
for (int i = t;i > 0;--i) {
ans[i] = d[i][y];
y = n[i] + y - d[i][y];
}
printf("%d\n",mi);
for (int i = 1;i <= t;++i)
printf("%d\n",ans[i]);
return 0;
}
第四题
#include <stdio.h>
#include <iostream>
#define MAXN 51
#define MAXL 10000000
using namespace std;
struct ss {
int x,y,dis,state;
} q[MAXL];
int n,m,tex,tey,x,y,state,dis,l,r;
int a[MAXN][MAXN];
bool v[65527][MAXN][MAXN];
int xx[4] = {1,-1,0,0};
int yy[4] = {0,0,1,-1};
void add(int x,int y,int di,int state) {
if ((x == n) && (y == n)) {
printf("%d\n",dis + 1);
exit(0);
}
if (a[x][y] < 0)
return;
if (a[x][y] == 0) {
r = (r + 1) % MAXL;
q[r].x = x;
q[r].y = y;
q[r].dis = di + 1;
q[r].state = state;
v[state][x][y] = 1;
return;
}
if (a[x][y] <= m) {
state |= (1 << (a[x][y] - 1));
r = (r + 1) % MAXL;
q[r].x = x;
q[r].y = y;
q[r].dis = di + 1;
q[r].state = state;
v[state][x][y] = 1;
return;
}
if ((state >> (a[x][y] - m - 1)) & 1) {
r = (r + 1) % MAXL;
q[r].x = x;
q[r].y = y;
q[r].dis = di + 1;
q[r].state = state;
v[state][x][y] = 1;
return;
}
}
bool inmap(int x,int y) {
return (x > 0) && (x <= n) && (y > 0) && (y <= n);
}
int main() {
freopen("syndicate.in","r",stdin);
freopen("syndicate.out","w",stdout);
scanf("%d%d",&n,&m);
for (int i = 1;i <= n;++i)
for (int j = 1;j <= n;++j)
scanf("%d",&a[i][j]);
for (int i = 1;i <= n;++i)
for (int j = 1;j <= n;++j)
if (a[i][j] == -2)
for (int k = 0;k < 4;++k) {
tex = i + xx[k];
tey = j + yy[k];
if (inmap(tex,tey) && (a[tex][tey] != -2))
a[tex][tey] = -1;
}
/*
for (int i = 1;i <= n;++i) {
for (int j = 1;j <= n;++j)
printf("%d ",a[i][j]);
printf("\n");
}
*/
q[0].x = q[0].y = 1;
while (l <= r) {
x = q[l].x;
y = q[l].y;
dis = q[l].dis;
state = q[l].state;
for (int i = 0;i < 4;++i) {
tex = x + xx[i];
tey = y + yy[i];
if (inmap(tex,tey) && (!v[state][tex][tey]))
add(tex,tey,dis,state);
}
l = (l + 1) % MAXL;
}
}