携程编程大赛 （预赛第二场）

A：和食物链做法一样，带权并查集

B：dp，01背包背出所有能组成边情况，在用这些情况去计算面积保留最大值

C：每个点从后往前搜，搜到合适就输出，搜不到就输出255 255 255

D：博弈，如果成对成对出现后手胜，否则先手胜

A:

#include <stdio.h>#include <string.h>const int N = 10005;int t, n, m, parent[N], rank[N];int find(int x) {    if (parent[x] == x)    return x;    int t = find(parent[x]);    rank[x] = (rank[parent[x]] + rank[x]) % 3;    return parent[x] = t;}int main() {    scanf("%d", &t);    while (t--) {    scanf("%d%d", &n, &m);    for (int i = 1; i <= n; i++) {        parent[i] = i;        rank[i] = 0;    }    int ans = 0;    while (m--) {        int q, x, y;        scanf("%d%d%d", &q, &x, &y);        if (x > n || y > n)        ans++;        else if (q == 2 && x == y) {        ans++;        }        else {        int px = find(x);        int py = find(y);        if (px == py) {            if (q == 1 && rank[x] != rank[y])            ans++;            else {            if (q == 2 && rank[y] != (rank[x] + 1) % 3)                ans++;            }        }        else {            parent[py] = px;            rank[py] = (rank[x] - rank[y] + 2 + q) % 3;        }        }    }    printf("%d\n", ans);    }    return 0;}

B:

#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>using namespace std;#define INF 0x3f3f3f3fconst int N = 805;int n, num[N], dp[N][N], sum;double cal(int a, int b, int c) {    double x = (a * a - b * b + c * c) * 1.0 / (2 * c * 1.0);    double h = sqrt(a * a * 1.0 - x * x);    return c * h / 2;}int main() {    while (~scanf("%d", &n) && n) {    sum = 0;    for (int i = 0; i < n; i++) {        scanf("%d", &num[i]);        sum += num[i];    }    memset(dp, 0, sizeof(dp));    dp[0][0] = 1;    for (int i = 0; i < n; i++) {        for (int j = sum / 2; j >= 0; j--) {        for (int k = sum / 2; k >= 0; k--) {            if (j - num[i] >= 0) {            if (dp[j - num[i]][k])                dp[j][k] = 1;            }            if (k - num[i] >= 0) {            if (dp[j][k - num[i]])                dp[j][k] = 1;            }        }        }    }    double ans = -INF;    for (int i = 0; i <= sum / 2; i++) {        for (int j = 0; j <= sum / 2; j++) {        if (dp[i][j] == 0) continue;        int k = sum - i - j;        if (i + j > k && k + j > i && i + k > j) {            ans = max(ans, cal(i, j, k));        }        }    }    if (ans < 0) printf("-1\n");    else printf("%d\n", (int)(ans * 100));    }    return 0;}

C:

#include <stdio.h>#include <string.h>const int N = 1005;int n, m;struct BIT {    int x1, y1, x2, y2, r, g, b;    void scanf_() {    scanf("%d%d%d%d%d%d%d", &x1, &y1, &x2, &y2, &r, &g, &b);    }} b[N];int main() {    while (~scanf("%d%d", &n, &m) && n + m) {    for (int i = 0; i < n; i++)        b[i].scanf_();    int x, y;    while (m--) {        scanf("%d%d", &x, &y);        int i;        for (i = n - 1; i >= 0; i--) {        if (x >= b[i].x1 && x <= b[i].x2 && y >= b[i].y1 && y <= b[i].y2) {            printf("%d %d %d\n", b[i].r, b[i].g, b[i].b);            break;        }        }        if (i == -1) printf("255 255 255\n");    }    }    return 0;}

D:

#include <stdio.h>#include <string.h>const int N = 15;int n, num, vis[105];bool judge() {    for (int i = 0; i <= 100; i++)    if (vis[i] % 2) return true;    return false;}int main() {    while (~scanf("%d", &n) && n) {    memset(vis, 0, sizeof(vis));    for (int i = 0; i < n; i++) {        scanf("%d", &num);        vis[num]++;    }    if (judge()) printf("Win\n");    else printf("Lose\n");    }    return 0;}

坚持.....