台州 OJ 2378 Tug of War

描述

A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight of the people on each team should be as nearly equal as possible.

输入

The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic.

输出

Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.

给 n 个人和他们各自的重量，要求把这 n 个人分成两部分，使这两部分的重量差值最小， 并且这两部分的人数差要小于等于 1。

用 vector 存放每个重量可能的人数，记得用一个 vis 数组判重，否则会超时。

代码：

#include <iostream>
#include <cstring>
#include <vector>
using namespace std;

const int MAX = 450*100/2 + 5;

vector<int> dp[MAX];    //到达 i 重量可能有多少人
bool vis[MAX][105];
int w[105];
int n;

int main(){
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);

    int tot = 0;
    cin >> n;
    for(int i=1; i<=n; i++){
        cin >> w[i];
        tot += w[i];
    }

    int maxSum = 0;
    memset(vis, 0, sizeof(vis));
    dp[0].push_back(0);        // 0 重量需要 0 个人
    vis[0][0] = 1;
    for(int i=1; i<=n; i++){
        for(int j=min(maxSum, tot/2); j>=0; j--){
            for(int k=0; k<dp[j].size(); k++){
                if(vis[j+w[i]][dp[j][k]+1] == 0){
                    dp[j+w[i]].push_back(dp[j][k] + 1);
                    vis[j+w[i]][dp[j][k]+1] = 1;
                }
            }
        }
        maxSum += w[i];
    }

    int ans = -1;
    for(int i=tot/2; i>=0; i--){
        for(int j=0; j<dp[i].size(); j++){
            if(n % 2 == 1){
                if(dp[i][j] >= n/2 && dp[i][j] <= n/2+1){
                    ans = i;
                    break;
                }
            }else{
                if(dp[i][j] == n/2){
                    ans = i;
                    break;
                }
            }
        }
        if(ans != -1)
            break;
    }

    cout << ans << " " << tot-ans;

    return 0;
}