題目鏈接：11578 - Situp Benches

題意：健♂身♂房有兩個仰臥起坐坐墊，每次調整角度要花費10元/10度，每次使用要花費15，如今給定n個人的時間順序，和所希望的角度，求最少花費 思路：dp，dp[i][j][k]表示第i個人，一個角度為j,還有一個為k的最小花費，一個人用和兩個人用的情況分開討論，然后記錄dp狀態(tài)轉移路徑。這個輸出路徑讓這題變得麻煩了不少。只是機智的我還是把它搞♂出♂來♂了。 代碼： #include <stdio.h> #include <string.h> #include <stdlib.h> #include <algorithm> using namespace std; #define INF 0x3f3f3f3f #define min(a,b) ((a)<(b)?(a):(b)) const int N = 10005; int t, n, i, j, k, dp[N][5][5], ans, an[N]; struct Stu {int t, l, id; } s[N];struct Out {int n, l, r, out1, out2; } out[N][5][5];bool cmpt(Stu a, Stu b) {return a.t < b.t; }bool cmpid(Stu a, Stu b) {return a.id < b.id; }void print(int n, int l, int r) {Out next = out[n][l][r];if (n == 0) return;if (next.out2 != -1) {an[s[n - 1].id] = next.out1;an[s[n].id] = next.out2;}else {an[s[n].id] = next.out1;}print(next.n, next.l, next.r); }int main() {scanf("%d", &t);while (t--) {ans = INF;memset(dp, INF, sizeof(dp));dp[0][0][0] = 0;scanf("%d", &n);for (i = 1; i <= n; i++) {scanf("%d%d", &s[i].t, &s[i].l);s[i].l = s[i].l / 10 - 1;s[i].id = i;}sort(s + 1, s + n + 1, cmpt);for (i = 1; i <= n; i++) {int tmp1 = s[i].l;if (i == n || s[i].t != s[i + 1].t) {for (j = 0; j < 5; j++) {for (k = 0; k < 5; k++) {if (dp[i][tmp1][k] > dp[i - 1][j][k] + abs(tmp1 - j) * 10) {dp[i][tmp1][k] = dp[i - 1][j][k] + abs(tmp1 - j) * 10;out[i][tmp1][k].l = j; out[i][tmp1][k].r = k; out[i][tmp1][k].n = i - 1;out[i][tmp1][k].out1 = 1; out[i][tmp1][k].out2 = -1;}if (dp[i][j][tmp1] > dp[i - 1][j][k] + abs(tmp1 - k) * 10) {dp[i][j][tmp1] = dp[i - 1][j][k] + abs(tmp1 - k) * 10;out[i][j][tmp1].l = j; out[i][j][tmp1].r = k; out[i][j][tmp1].n = i - 1;out[i][j][tmp1].out1 = 2; out[i][j][tmp1].out2 = -1;}}}}else {int tmp2 = s[i + 1].l;for (j = 0; j < 5; j++) {for (k = 0; k < 5; k++) {if (dp[i + 1][tmp1][tmp2] > dp[i - 1][j][k] + abs(tmp1 - j) * 10 + abs(tmp2 - k) * 10) {dp[i + 1][tmp1][tmp2] = dp[i - 1][j][k] + abs(tmp1 - j) * 10 + abs(tmp2 - k) * 10;out[i + 1][tmp1][tmp2].l = j; out[i + 1][tmp1][tmp2].r = k; out[i + 1][tmp1][tmp2].n = i - 1;out[i + 1][tmp1][tmp2].out1 = 1; out[i + 1][tmp1][tmp2].out2 = 2;}if (dp[i + 1][tmp2][tmp1] > dp[i - 1][j][k] + abs(tmp2 - j) * 10 + abs(tmp1 - k) * 10) {dp[i + 1][tmp2][tmp1] = dp[i - 1][j][k] + abs(tmp2 - j) * 10 + abs(tmp1 - k) * 10;out[i + 1][tmp2][tmp1].l = j; out[i + 1][tmp2][tmp1].r = k; out[i + 1][tmp2][tmp1].n = i - 1;out[i + 1][tmp2][tmp1].out1 = 2; out[i + 1][tmp2][tmp1].out2 = 1;}}}i++;}}int lv, rv;for (j = 0; j < 5; j++) {for (k = 0; k < 5; k++) {if (ans > dp[n][j][k] + j * 10 + k * 10) {ans = dp[n][j][k] + j * 10 + k * 10;lv = j; rv = k;}}}printf("%d\n", ans + 15 * n);print(n, lv, rv);for (i = 1; i <= n; i++)printf("%d\n", an[i]);}return 0; }

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