wenbao与cf（849D）思维

http://codeforces.com/contest/848/problem/B

考虑会碰到的条件，X+Y相等，贪心

#include <iostream>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn = 100009;
int g[maxn], p[maxn], t[maxn], n, w, h, x[maxn], y[maxn], co_x[maxn], co_y[maxn];
vector<int> v[maxn*2];

bool cmp(int x, int y){
    if(g[x] != g[y]) return g[x] == 2;
    else{
        if(g[x] == 1) return p[x] < p[y];
        else return p[x] > p[y];
    }
}

int main(){
    scanf("%d%d%d", &n, &w, &h);
    for(int i = 0; i < n; ++i){
        scanf("%d%d%d", &g[i], &p[i], &t[i]);
        v[p[i]-t[i]+maxn].push_back(i);
    }
    for(int i = 0; i < maxn*2; ++i){
        int numx = 0, numy = 0;
        for(int j = 0; j < v[i].size(); ++j){
            int xx = v[i][j];
            if(g[xx] == 1) x[numx++] = p[xx];
            else y[numy++] = p[xx];
        }
        sort(x, x+numx);
        sort(y, y+numy);
        sort(v[i].begin(), v[i].end(), cmp);
        for(int j = 0; j < numx; ++j){
            co_x[v[i][j]] = x[j];
            co_y[v[i][j]] = h;
        }
        for(int j = 0; j < numy; ++j){
            co_x[v[i][numx+j]] = w;
            co_y[v[i][numx+j]] = y[numy-j-1];
        }
    }
    for(int i = 0; i < n; ++i){
        printf("%d %d
", co_x[i], co_y[i]);
    }
    return 0;
}

 看到别人博客里面提供了另外一种思路，可以将起点和终点排序必然是一一对应的

#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 100009;
struct Node{
    int x, y, xx, yy, id;
};

Node a[maxn], b[maxn];
int co_x[maxn], co_y[maxn];

int main(){
    int n, w, h, g, p, t;
    scanf("%d%d%d", &n, &w, &h);
    for(int i = 0; i < n; ++i){
        scanf("%d%d%d", &g, &p, &t);
        if(g == 1){
            a[i] = (Node){p, -t, p, h, i};
        }else{
            a[i] = (Node){-t, p, w, p, i};
        }
        b[i] = a[i];
    }
    //N1.x+N1.y == N2.x+N2.y 可以相撞
    sort(a, a+n, [](Node N1, Node N2){
        return (N1.x+N1.y < N2.x+N2.y || N1.x+N1.y == N2.x+N2.y && N1.x-N1.y < N2.x-N2.y);
    });
    sort(b, b+n, [](Node N1, Node N2){
        return (N1.x+N1.y < N2.x+N2.y || N1.x+N1.y == N2.x+N2.y && N1.xx-N1.yy < N2.xx-N2.yy);
    });
    for(int i = 0; i < n; ++i){
        co_x[a[i].id] = b[i].xx;
        co_y[a[i].id] = b[i].yy;
    }
    for(int i = 0; i < n; ++i){
        printf("%d %d
", co_x[i], co_y[i]);
    }
    return 0;
}

只有不断学习才能进步！