UVA-1609

這個題就是純構造，一旦找到“黑隊”和“白隊”的概念，就有一定的思路了。
後面就是構造四個階段。每一個階段的原則就是儘量讓白隊打黑隊，然後讓黑隊內部打，最後隨機分配。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1024+5;
char table[maxn][maxn];

int main(void){
    int n;
    while(scanf("%d",&n)==1){
        vector <int> win,lose;
        for(int i=1;i<=n;i++){
            scanf("%s",table[i]+1);
        }
        for(int j=2;j<=n;j++){
            if(table[1][j]=='1') win.push_back(j);
            else lose.push_back(j);
        }
        //win代表1可以y贏，lose代表1會輸。
        //lose是黑子

        int ct = n;
        while(ct>1){
            vector<int> win2,lose2,nextt;

            for(int i=0;i<lose.size();i++){
                int tlose=lose[i];
                bool match=false;
                for(int j=0;j<win.size();j++){
                    if(win[j]==0) continue;
                    int twin=win[j];
                    if(table[twin][tlose]=='1'){
                        printf("%d %d\n",twin,tlose);
                        win2.push_back(twin);
                        win[j]=0;
                        match=true;
                        break;
                    }
                }
                if(!match) nextt.push_back(tlose);
            }

            bool first=true;
            for(int i=0;i<win.size();i++){
                int twin=win[i];
                if(win[i]){
                    if(first){
                        first=false;
                        printf("1 %d\n",win[i]);
                    }
                    else{
                        nextt.push_back(twin);
                    }
                }
            }


            for(int i=0;i<nextt.size();i+=2){
                printf("%d %d\n",nextt[i],nextt[i+1]);
                int keep = nextt[i];
                if(table[nextt[i+1]][keep]=='1') keep=nextt[i+1];
                if(table[1][keep]=='1') win2.push_back(keep);
                else lose2.push_back(keep);
            }


            win=win2;
            lose=lose2;
            ct>>=1;
        }
    }
    return 0;
}