2017百度之星资格赛 1003 度度熊与邪恶大魔王(DP)

题目链接：点击打开链接

题目中文，容易理解，不再叙述

思路：

开始想到的是完全背包，对于每个怪兽，分别去利用完全背包得到最后DP结果，可是怪兽数量（取值1~10^5）太大，复杂度达到O(10^9)，所以一定会超时，而且思路不是最佳的

正确的解法就是普通的DP，dp[i][t]表示防御力为i（取值0~10），生命值为t（取值1~1000）的怪兽被打败需要的最小晶石数，再加入j（取值1~1000）表示技能，直接枚举i，j，t（其中i，j要满足harm[j] >= i），由状态转移方程得最后结果，然后再去处理每个怪兽。复杂度为O(10^7)

// 度度熊与邪恶大魔王.cpp 运行/限制：202ms/1000ms#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;#define INF 0x3f3f3f3fint stone[1005], harm[1005];int life[100005], pro[100005];int dp[15][1005];int main(){int n, m;while (scanf("%d%d", &n, &m) != EOF) {for (int i = 0; i < n; i++) {scanf("%d%d", &life[i], &pro[i]);}for (int i = 0; i < m; i++) {scanf("%d%d", &stone[i], &harm[i]);}memset(dp, INF, sizeof(dp));for (int i = 0; i <= 10; i++) {//防御力for (int j = 0; j < m; j++) {//技能if (harm[j] < i) continue;for (int t = 1; t <= 1000; t++) {//生命力if (harm[j] - i >= t) dp[i][t] = min(dp[i][t], stone[j]);else dp[i][t] = min(dp[i][t], dp[i][t - (harm[j] - i)] + stone[j]);}}}long long re = 0;int flag = 1;for (int i = 0; i < n; i++) {if (dp[pro[i]][life[i]] == INF) {flag = 0;break;}re += dp[pro[i]][life[i]];}if (flag) printf("%lld\n", re);else printf("-1\n");}    return 0;}

TLE的完全背包代码：

// 度度熊与邪恶大魔王.cpp Time Limit Exceeded#include "stdafx.h"#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int life[100005], pro[100005];int stone[1005], harm[1005];int dp[1005];#define INF 0x3f3f3f3fint main(){int n, m,re, flag;while (scanf("%d%d", &n, &m) != EOF) {for (int i = 0; i < n; i++) {scanf("%d%d", &life[i], &pro[i]);}for (int i = 0; i < m; i++) {scanf("%d%d", &stone[i], &harm[i]);}re = 0, flag = 1;for (int i = 0; i < n; i++) {//怪兽 10^5memset(dp, INF, sizeof(dp));for (int j = 0; j < m; j++) {//技能 10^1if (harm[j] < pro[i]) continue;for (int t = 1; t <= life[i]; t++) {//生命值 10^3if (harm[j] - pro[i] >= t) {dp[t] = min(dp[t], stone[j]);}else {dp[t] = min(dp[t], dp[t - (harm[j] - pro[i])] + stone[j]);}}}if (dp[life[i]] == INF) {printf("-1\n");flag = 0;break;}re += dp[life[i]];}if (flag) {printf("%d\n", re);}else {printf("-1\n");}}    return 0;}