uva 10895 - Matrix Transpose(STL)

题目链接：uva 10895 - Matrix Transpose

题目大意：给定一个矩阵，求它的转置。矩阵会忽略元素为0的位置。

解题思路：矩阵aij会跑到aji的位置，模拟一下接好了。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 1e4+5;

struct item {
    int pos, val;
    item (int pos = 0, int val = 0) {
        this->pos = pos;
        this->val = val;
    }
};

bool cmp (const item& a, const item& b) {
    return a.pos < b.pos;
}

int N, M, arr[105];
vector<item> g[maxn];

int main () {
    while (scanf("%d%d", &N, &M) == 2) {
        for (int i = 0; i <= M; i++)
            g[i].clear();

        int k, x;
        for (int i = 1; i <= N; i++) {
            scanf("%d", &k);
            for (int j = 0; j < k; j++)
                scanf("%d", &arr[j]);

            for (int j = 0; j < k; j++) {
                scanf("%d", &x);
                g[arr[j]].push_back(item(i, x));
            }
        }

        printf("%d %d\n", M, N);
        for (int i = 1; i <= M; i++) {
            sort(g[i].begin(), g[i].end(), cmp);

            printf("%lu", g[i].size());
            for (int j = 0; j < g[i].size(); j++)
                printf(" %d", g[i][j].pos);
            printf("\n");

            if (g[i].size() != 0) {
                printf("%d", g[i][0].val);
                for (int j = 1; j < g[i].size(); j++)
                    printf(" %d", g[i][j].val);
            }
            printf("\n");
        }
    }
    return 0;
}