2015 Asia - Jakarta D - An ICPC Problem without Statement

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题目描述：

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=104601#problem/D

题解：

1.当一定大于等于0的时候,就是我们尽量凑2的时候,我们放2,-2,1, -1, 0的顺序,每次能放就放最多的,和-1的 dfs深搜.
2.当必须是-的时候,要么全部必须要选,要么全是负数(可以推一下) 所以就是绝对值小的.

重点：

1.负数的时候单拉出来绝对值最小.
2.减一的小方法

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;const int maxn = 1e5 + 100;int n, A, B;int a[maxn];int ans[maxn], val, tag;int b[20], number[10], have[20];int num;int change(int x) {    if(x == 2)    return 0;    if(x == -2)    return 1;    if(x == 1)    return 2;    if(x == -1)    return 3;    return 4;}void dfs(int pos, int now) {    if(pos == 4) {    if(now + have[number[pos]] < num)        return;    int t = num - now;    if(t == 0) {        int tval = b[0] + b[1];        int ttag = (b[1] + b[3]) % 2;        if(ttag == 1)        ttag = -1;        else        ttag = 1;        if(ttag > tag || (ttag == tag && (tag == 1 && tval > val || tag == -1 && tval < val)))        {            tag = ttag;            val = tval;            for(int i = 0; i <= 3; i++) {            ans[i] = b[i];            }            ans[4] = num - now;        }    }    else {        if(tag < 0) {        tag = 0;        val = 0;        for(int i = 0; i <= 3; i++) {            ans[i] = b[i];        }        ans[4] = num - now;        }    }    return;    }    int tmp = min(num - now, have[number[pos]]);    b[pos] = tmp;    dfs(pos + 1, now + tmp);    if(tmp > 0) {    b[pos] = tmp - 1;    dfs(pos + 1, now + tmp - 1);    }}vector<int> res;void solve() {    tag = -2;    for(num = A; num <= B; num++)    dfs(0, 0);    /*printf("%d  %d\n", tag, val);      for(int i = 0; i <= 4; i++) {      printf("%d ", ans[i]);      }*/    res.clear();    if(tag >= 0) {    ;    }    else {    if(A == n && A == B) {        for(int i = 0; i <= 4; i++)        ans[i] = have[i];    }    else {        for(int i = 0; i <= 4; i++)        ans[i] = 0;        ans[3] = min(A, have[3]);        ans[1] = A - ans[3];    }    }    for(int i = 0; i < n; i++) {    if(ans[change(a[i])] > 0) {        res.push_back(i);        ans[change(a[i])]--;    }    }    printf("%d\n", res.size());    for(int i = 0; i < res.size(); i++) {    printf("%d%c", res[i] + 1, (i == res.size() - 1 ? '\n' : ' '));    }}int main() {    //freopen("d.txt", "r" ,stdin);    number[0] = 0;    number[1] = 1;    number[2] = 2;    number[3] = 3;    number[4] = 4;    int ncase;    scanf("%d", &ncase);    for(int _ = 1; _ <= ncase; _++) {    printf("Case #%d:\n", _);    scanf("%d %d %d", &n, &A, &B);    for(int i = 0; i <= 4; i++)        have[i] = 0;    for(int i = 0; i < n; i++) {        scanf("%d", &a[i]);        have[change(a[i])]++;    }    solve();    }    return 0;}